Introduction:
This is my first answer on Stackexchange so feel free to correct any mistakes I will undoubtedly make, especially with the math formatting as I am not very good at that either(although I will try my best). I also know that I am a bit late to answer this question, but I hope that this answer will still bring some insight into why moving the Moon is so hard.
I will split this into 2 parts, for the 2 ways I can think of to move the Moon. Part 1 will address the method of hitting the moon with an asteroid to change its speed and part 2 will look at continuous thrust. I will use the same type of setup to Benjamin's answer with an unstoppable asteroid that will hit the Earth in 250 years, but if we can move the Moon in such a way that it is half an orbit early or late we can stop it(The probability of the killer asteroid to be in the same plane as the Moon is very low but I will ignore that small technicality for now).
In both parts I will assume the worst case scenario with the Moon being exactly 180 degrees off where it needs to be. At the end of both parts I will provide a distilled equation for any deadline time other than 250 years, so that you can plug in any year and get out the expected resources that you will need. There will also be a tldr at the very end for those that get horrified by all the math.
Method 1: Guided asteroid impact
So the goal of this part is to find the mass $M_a$ of the asteroid that we send to the Moon. To do that, we first need to figure out how much speed the Moon will lose during the collision. The current speed of the Moon can be figured out using the equation $\sqrt{\frac{\mu}{a_m}}$ where $a_m$ is the radius of the Moons orbit(usually called the semi-major axis, That will be important later when talking about the non-circular orbit of the asteroid) and $\mu$ is the standard gravitational parameter for the Earth and the Moon. Plugging in the values we get:
$a_m=384399000m$
$\mu=4.034713308×10^{14}\frac{m^3}{s^2}$
$v_m=\sqrt{\frac{\mu}{a_m}}=1024.5\frac{m}{s}$
Now when we know the speed of the Moon, we need to know the speed of the asteroid, and That requires knowing the orbit of the asteroid. I will pick an orbit that goes between the asteroid belt and the Earth(a fairly typical path for a near Earth asteroid). So I will pick a perihelion(inner point) of 1 au(Earths orbit), and an aphelion(outer point) at 3 au(in the middle of the asteroid belt). So we have $p_0=1$ and $p_1=3$. Now we need the speed of the asteroid when it is arrives at Earth(when it is at $p_0$), but the simple speed-from-radius equation we used earlier is not going to work this time as the orbit is elliptic instead of circular. So we need an upgraded version called the vis-viva equation:
$v=\sqrt{\mu\left(\frac{2}{r}-\frac{1}{a}\right)}$
Because I am using units where Earths speed and Earths orbital radius are all set to 1, That $\mu$ term is also 1(note that as both Earth and the asteroid are orbiting around the Sun, That \mu is for the Sun, it is not the same value as earlier). The $r$ and $a$ terms are the current radius and semi-major axis of the orbit respectively. The $r$ we know as the asteroids current position is Earths orbit at $p_0$, the semi-major axis $a$ is simply the average between $p_0$ and $p_1$, ie 2. We also want to multiply the whole thing with the velocity of Earth $v_e=29780 \frac{m}{s}$ to get out of the whole Earth=1 thing and back to normal units. Plugging it all in gives:
$v_a=v_e\sqrt{\frac{2}{1}-\frac{1}{2}}=36472.9\frac{m}{s}$
To get the maximum impact velocity we want the asteroid to hit the Moon when the Moon goes head on to the asteroid from Earths perspective. This gives:
$v_i=\left(v_a-v_e\right)+v_m=7717.4\frac{m}{s}$
So now we know how fast the asteroid will hit the Moon, unfortunately here we need the mass of the asteroid, and we don't have that yet. So from here on I will skip the number crunching, and use only variables.
In any collision, momentum is conserved. I will use this to figure out how much the Moon will slow down when struck by our asteroid. The momentum of an object is $mass\cdot velocity$, so the momentum of the Moon before the collision is $M_mv_m$ where $M_m=7.342\times10^{22}kg$ is the mass of the Moon. The momentum of the asteroid is $M_av_i$. The momentum of the Moon after the collision is $M_mv_m-M_av_i$. To get the new velocity we simply divide by the mass of the Moon:
$v_{m2}=\frac{M_mv_m-M_av_i}{M_m}$
To be able to get the orbital period that we will need later we need to get the new semi-major axis of the Moon after the collision. To do that we use the vis-viva equation again, but in reverse. Things are abit different this time because the current radius $r$ is actually the old semi-major axis $a_m$, and we need to use the $\mu$ this time Because we are orbiting around the Earth. Lets plug things in:
$v_{m2}=\sqrt{\mu\left(\frac{2}{a_m}-\frac{1}{a_{m2}}\right)}$
And solve for $a_{m2}$ to get:
$a_{m2}=\frac{\mu a_m}{2\mu-a_m\left(v_{m2}\right)^2}$
So, remember that I mentioned orbital period? Here is where that comes in. The equation for orbital period is:
$P\left(a\right)=\tau\sqrt{\frac{a^3}{\mu}}$
It takes in the semi-major axis $a$ and spits out the amount of seconds it takes to go one lap around the Earth at that height. We need to use that function in order to construct our final equation for how massive of an asteroid you need to smack the Moon with, so that it can catch up to and be smacked by a much larger asteroid in 250 years. Poor Moon...
$\left(P\left(a_m\right)-P\left(a_{m2}\right)\right)\cdot250\cdot12=\frac{P\left(a_m\right)}{2}$
So here it is. It says that we want the change in time for one orbit times the number of orbits to be equal to half of the time it takes to make one orbit. So when the $M_a$ is just right and this equation is true, the Moon will travel half an orbit early after 250 years and catch the larger asteroid. I will spare you the walk through of solving this, but if you feel brave enough you can try it yourself.
$M_a=\frac{M_m}{v_i}\left(v_m-\sqrt{\frac{u}{a_m}\left(2-\left(1-\frac{1}{2\cdot\left(Y\cdot12\right)}\right)^{-\frac{2}{3}}\right)}\right)$
What a beast! But it does its job perfectly. If i plug in $Y=250$ it spits out $M_a=5.41×10^{17}kg$. That is alot of kilograms. We can also plug that into this formula to get the diameter of the asteroid: $d=2\left(\frac{M_a}{3000}\frac{3}{4\pi}\right)^{\frac{1}{3}}=70km$. 70km is a large asteroid, not many asteroids in the asteroid belt are that large(about 400 or so). Although both the mass and diameter is completely dwarfed by the Moon(Moon is 135567 times more massive and 50 times as wide), and this explains why moving the Moon is hard even when you have 250 years to do it.
Method 2: Continuous Thrust
This part focuses on how much thrust is needed to move the Moon by the same amount as in Part 1, if the thrust is applied continuously over 250 years. Luckily for use the approach is quite similar at the start and we will see many of the same variables from the first part, so I won't explain them twice. So if you see $\mu$ for example, you know that it is the exactly same thing as before.
With that out of the way, lets start with the plan. This time I will use kinetic energy as the conserved quantity instead of momentum. That is because $force\cdot time=energy$, so the decrease in kinetic energy is proportional to time:
$K\left(t\right)=\frac{1}{2}m_mv_m^2-Ft$
As You can see, $K\left(t\right)$ is a function of time, and the only new variable is the thruster force $F$. Yep, that crucial variable that we need to get our hands on at the end of this whole chapter is right there in the very first equation, and that means that I won't do any number crunching until the end. Let's continue in an similar style of "conserved quantity -> $v_{m2}$ -> $a_{m2}$ -> $P(a)$ -> result" that we used in Part 1. But because our conserved quantity is a function of time, so will all of these values be functions of time in this part. Anyways, lets go on to finding $v_{m2}\left(t\right)$. That can be easily solved by reverse engineering the kinetic energy equation to get:
$v_{m2}\left(t\right)=\sqrt{\frac{2K\left(t\right)}{m_m}}$
The equation for velocity to semi-major axis is exactly the same as before, except that it is a function of time, because $v_{m2}\left(t\right)$ is a function of time:
$a_{m2}\left(t\right)=\frac{ua_m}{2u-a_m\left(v_{m2}\left(t\right)\right)^2}$
The period function $P\left(a\right)$ is completely identical, so I won’t bother to show it again. This means that we are already at the point where I show the final equation that puts the constraint that "the difference in each orbit times the amount of orbits should equal half of the original orbit", But hang on a sec... That "difference in each orbit" part is now a function of time, every single second it will be different. Therefore we will need some heavy duty mathematical machinery, namely the (in)famous integral operator from calculus, I know that it sounds scary, but I will both explain it and (painfully) solve it. Anyways, here we go...
$\int_0^T\left(P\left(a_m\right)-P\left(a_{m2}\left(t\right)\right)\right)dt=\frac{P\left(a_m\right)}{2}$
What a monster, isn't it? But if you look closely, You an see that it somewhat resembles the final equation from Part 1, the $P\left(a_m\right)$ and $\frac{P\left(a_m\right)}{2}$ parts are still there, doing the same job as they did before. The $P\left(a_{m2}\right)$ has been upgraded to $P\left(a_{m2}\left(t\right)\right)$ to include time, and the factor of $250\cdot 12$ has been replaced by that scary looking $\int$ integral thing. That integral is there to take care of the fact that during the whole operation from $0$ seconds to $T$ seconds, the value inside the "body" of the integral is constantly changing. I won't go into details as to exactly how the integral manages to do this, rather I will skip the (long) time it took for me to find a solution to this and actually show the solution itself. Oh and by the way, that $T$ is just the amount of seconds there is in 250 years(big number, I know), it is equal to $T=31536000\cdot Y$ where the $Y$ is the same variable for years that we saw earlier. Ok, here is the painfully acquired solution to that scary integral formula thing:
$F=\left(\frac{m_mu}{a_m}\right)\frac{3+\frac{4}{2T-1}-\sqrt{\frac{18T-1}{2T-1}}}{4T}$
That almost looks even scarier than the integral itself, luckily this is just a rather large normal "plug and play" function, where You just plug in all the values as they are and it will spit out the awnser. For the value of $T$ that we need for it to represent 250 years, the formula spits out $F=413266741N$ that is 412 MN(MegaNewtons) or roughly equal to 53 Saturn-V first stage engines, that is 10 copies of the most powerful rocket ever made pushing the Moon for 250 years straight… That approach is also very hard.
Conclusion/tldr:
The Moon is big, very big, and heavy. So pushing it around is not something you can do as if it was any regular old asteroid.
tldr for Method 1: It would take a 70 km asteroid head on to the Moon to be anywhere close to making it in time, that is much larger than the majority of asteroids in the asteroid belt and is about the same size as the asteroid that killed the dinosaurs(although the size of that one is not very well-known).
tldr for Method 2: In order to move the Moon using regular rocket engines,it would take 10 copies of Saturn-V(the rocket that took people to the Moon), blasting away at the Moon for 250 years straight, so that is probably not doable either.