The short answer is that it is indeed possible to create systems with negative temperatures. Unfortunately, that doesn't imply that the system in question has a negative energy density.
What is temperature?
I think you need to move away from the idea of temperature as simply a measure of the kinetic energy per particle in a system. Other folks have used it, but in thermodynamics, it can be extremely misleading - temperature encapsulates a much wider range of behaviors than just moving around. Instead, I like to use the statistical mechanics definition of temperature, and think about it as a way of measuring how the number of possible configurations (called "microstates") of a system changes as you increase or decrease the system's energy.
Let's say we have a system consisting of two particles, with a total energy of two "chunks" of energy.$^{\dagger}$ How many ways can we distribute those chunks of energy between the particles?
| Energy of particle 1 |
Energy of particle 2 |
| 2 |
0 |
| 1 |
1 |
| 0 |
2 |
This gives us three possible ways we can distribute the energy; we then say that the system has three possible microstates. Now say we increase the energy of the system, and add a third chunk of energy. Now how many ways can the energy be distributed?
| Energy of particle 1 |
Energy of particle 2 |
| 3 |
0 |
| 2 |
1 |
| 1 |
2 |
| 0 |
3 |
We can distribute the energy four ways, so the system now has four possible microstates. By increasing the system's energy, it turns out that we've increased its number of possible microstates.
The entropy of a system, $S$, is related to the number of microstates $\Omega$ by the relation
$$S=k_B\ln\Omega$$
where $k_B$ is Boltzmann's constant. An increase in microstates means an increase in entropy. Here's where temperature comes in. The statistical mechanical way of defining temperature is
$$\frac{1}{T}=\frac{\partial S}{\partial E}$$
where $E$ is the energy of the system; put in words, temperature describes how the entropy of a system changes in response to changes in the system's energy.
What about negative temperature?
We can see that the toy system above has a positive temperature: An increase in the energy increases the entropy, and a decrease in the energy decreases the entropy. Therefore, for any change of energy, we always have the condition that
$$\frac{\partial S}{\partial E}>0$$
and so the system has a positive temperature. By using the general properties of Einstein solids, you can show, in fact, that regardless of the number of particles or the total amount of energy, the temperature of Einstein solid is always positive.
Now, negative temperature systems do exist, as you've alluded to in comments (e.g. Braun et al. 2013, which you mentioned above). However, they have some strange properties. For instance, the Hamiltonian describing them must be bounded from above (feel free to skip this section if you're not familiar with Hamiltonians - you won't miss too much!). This means we can show easily that certain systems can never have negative temperatures.
For example, consider an ideal gas of $N$ identical particles, each with mass $m$. We assume that the particles don't interact, so there is no potential term in the Hamiltonian - just a bunch of kinetic terms:
$$H=\sum_{i=1}^N\frac{\mathbf{p}_i^2}{2m}$$
where $\mathbf{p}_i$ is the momentum vector of particle $i$. As there is no limit to the momentum of a particle, the Hamiltonian is not bounded from above. Therefore, an ideal gas must have a positive temperature.
Now let's look at the Einstein solid case in more detail. Each "particle" is actually a three-dimensional quantum harmonic oscillator, with mass $m$, frequency $\omega$ and Hamiltonian
$$\hat{H}=\sum_{i=1}^N\frac{\hat{\mathbf{p}}_i^2}{2m}+\frac{1}{2}m\omega^2|\hat{\mathbf{r}_i}|^2$$
Again, the momentum operator is unbounded, so the Einstein solid must have a positive temperature. In general, unless the kinetic energy is somehow bounded, the presence of a kinetic term implies a positive temperature (as I understand it). To create a negative temperature system, you would need the appropriate bounds on the Hamiltonian.
In negative-temperature systems, if you increase the total energy, you decrease the entropy - and by our statistical definition of temperature, the system must have a negative temperature. This doesn't mean that the average energy per particle in the system has a negative energy; therefore, it's not useful for creating something like an Alcubierre drive.
There are some interesting consequences to this. In the absence of performing work (as is the case in a refrigerator, where work is performed to cool your food), when two objects with positive temperatures are placed in contact, heat will flow from the one with the higher temperature to the one with the lower temperature. On the other hand, if one object has a negative temperature and the other has a positive temperature, heat will always flow from the negative-temperature object to the positive-temperature object. In this sense, an object with negative temperature is always hotter than an object with positive temperature.
A way of quantifying this is to define
$$\beta\equiv\frac{1}{k_BT}=\frac{1}{k_B}\frac{\partial S}{\partial E}$$
and note that for any two objects with $\beta_1$ and $\beta_2$, heat will flow from object 1 to object 2 if $\beta_1<\beta_2$, and from object 2 to object 1 if $\beta_2<\beta_1$.
As I've stated above, we need to separate the notion of temperature from the notion of kinetic energy. In the classical case where we have, say, particles at a mean temperature $T$ moving around, then we can assign them each thermal energies $f\cdot\frac{1}{2}k_BT$, with $f$ the degrees of freedom of the particle. From this, you'd think that a negative temperature corresponds to a negative kinetic energy - but as the sort of negative-temperature systems we're talking about don't involve billiard ball-like particles moving around classically, it's not really correct to say that they have negative energies.
$^{\dagger}$This is effectively a simplistic Einstein solid model. In general, if there are $q$ chunks of energy and $N$ particles, there are
$$\Omega(q,N)=\frac{(q+N-1)!}{q!(N-1)!}$$
microstates. You can check that the two cases I described above correspond to $\Omega(2,2)=3$ and $\Omega(3,2)=4$.
