I am going to assume that by constant, you really mean constant as opposed to instantaneous. In other words, we are still bound by the speed of light propagation delay. We are also bound by the laws of orbital mechanics as currently understood.
Since you use "planets" plural, I take it that humanity has colonies on multiple planets and possibly moons, as opposed to just one outpost away from Earth or Earth-centric orbit (which we already have one of: the International Space Station).
I'm also going to, for simplicity's sake, assume that you have unlimited power output for the transmitters. In practice this is not going to be the case, but to a first order approximation to maintain reader suspension of disbelief, it works okay. Also, you can trade power output for data rate, as described by the Shannon–Hartley theorem, so if you can accept a lower data transmission rate, you can make do with less power (to a point).
Let's start with colonies only on the planets' surface, not any of the moons in the solar system. The problem here is that planets orbit the Sun with little regard to their respective orbital alignment with the other planets.
The easiest way to ensure that every planet is always in view of at least one communications relay satellite is probably to put the relay sats in an orbit around the Sun which is highly inclined relative to the solar system ecliptic (the imaginary disk that is formed by the orbits of the planets, which traces back to the protoplanetary disk of the solar system). A simple way to do that (well, "simple", but still expensive in terms of orbital maneuvering to get into place) would be to use a solar polar orbit. This is an orbit that goes across the poles of the Sun, rather than around the Sun's equator, at a 90 degree angle to the ecliptic.
Having three relay satellites in a solar polar orbit, 120 degrees out of phase, will ensure that there is always one within view of somewhere on every planet in the solar system, as the Sun will only block the view of one at any one time (as viewed from any particular planet). You may want a few extra for redundancy, but doing so does not significantly change the setup. Given that the other end of the link is close to the ecliptic, having three ensures that one is always within view of each planet, whereas with two, the situation could arise that one is behind the Sun and the other is directly in front of the Sun. That would almost certainly work from a geometric point of view, but in practice you would have serious trouble picking the signal out from the Sun's noise (see below).
Now, notice that I said somewhere on every planet. You are going to need a similar constellation in orbit around each planet where there is a human colony, to ensure that there is a satellite in view of every point on the surface where it is needed. At this point it comes down to a similar scenario to that described in Minimum number of satellites to image the entirety of Earth's surface at all times. It turns out that this is possible to do with four to six satellites (mostly depending on your ground station capabilities, I suppose; four is the absolute minimum required for the satellite constellation to be able to see every point on the surface all the time, but you also need particular spots on the surface to be able to communicate with at least one of the satellites at any given time). Again, you may want a few extra for redundancy, but solving this problem is not an insurmountable task.
Once you add colonies on the planets' moons or otherwise in orbit of the planets, you will need a reliable method for communication from the colony to the relay sats around the planet. For this, you can look to the Tracking and Data Relay Satellite System (TDRSS) for inspiration. Long story short, you need at least three satellites in geostationary orbit to maintain constant communications between any point in orbit and any point on the ground, after which getting the signal to the solar system relay sats is merely a matter of getting a signal (any signal) from point A to point B on the planet's or moon's surface, or between the TDRSS-like satellites. Once the signal is within view of one of the solar relay sats, have the satellite shoot it off toward the solar relay satellite, and the signal is off to its penultimate destination.
There are two big problems with this that your world's engineers would be facing, which I can think of.
First, the Sun is rather noisy well into the RF spectrum. That's a problem when the Sun is in line with the desired signal. So you will need to either place the Sun-orbiting satellites in a relatively high orbit around the Sun to ensure sufficient separation that high-gain antennas can select against the radio noise of the Sun, or extremely high-gain antennas at the ends of the links. I don't know which of these would be easier, but given that fighting the ecleptic is already difficult, either might well be worth the price to pay. Note that the higher-gain antennas require more accurate aiming, which will require more station-keeping, necessiting more reaction mass ("fuel") on-board the satellites for a given service life. Again, not insurmountable, but worth keeping in mind as it is an issue that real-life engineers would have to contend with and make tradeoffs in.
Second, solar polar orbit is hard. I alluded to this above, but don't dismiss the importance of it; it really is crazy hard. Let's say you want to place the Sun-orbiting relay satellites at the distance to the Sun of Venus (0.73 AU), with an inclination to the ecliptic of 90 degrees. First, you need to get to Venus' orbit, which can be accomplished with a Hohmann transfer (calculated based on a heliocentric, or Sun-centered, reference frame):
$$ r_1 = 1.00~\text{AU} \approx 149\,598\,023\,000~\text{m} \\
r_2 = 0.73~\text{AU} \approx 109\,206\,445\,611~\text{m} \\
\Delta v_1 = \sqrt{\frac{\mu_\text{Sun}}{r_1}} \left( \sqrt{\frac{2r_2}{r_1 + r_2}} - 1 \right) = \sqrt{\frac{1.3271244 \times 10^{20}}{149\,598\,023\,000}} \left( \sqrt{\frac{218\,412\,891\,222}{258\,804\,468\,611}} - 1 \right) \\
\Delta v_2 = \sqrt{\frac{\mu_\text{Sun}}{r_2}} \left( 1 - \sqrt{\frac{2r_1}{r_1 + r_2}} \right) = \sqrt{\frac{1.3271244 \times 10^{20}}{109\,206\,445\,611}} \left( 1 - \sqrt{\frac{299\,196\,046\,000}{258\,804\,468\,611}} \right) \\
\Delta v_1 \approx 2\,423~\text{m/s} \\
\Delta v_2 \approx 2\,622~\text{m/s} \\
\Delta v = \Delta v_1 + \Delta v_2 \approx 5\,045~\text{m/s} $$
which is managable (going to the Moon took a total of about 11 km/s delta-v for the trip out, plus some for landing and going back for a total delta-v budget of somewhere in the vicinity of 20 km/s split among the Saturn, service module, lunar module descent and lunar module ascent stages). This puts you in the neighborhood of Venus; not necessarily in Venus' actual location (that depends on orbital transfer timing, or what we refer to as launch windows), but at least approximately co-orbiting with it. Now, assume that your orbit is circular, and change its inclination by 90 degrees while maintaining its circularity (technically, its eccentricity), where $v = 35.02~\text{km/s}$ is Venus' orbital velocity around the Sun:
$$ \Delta v_i = 2v \sin\left({\frac{\Delta i}{2}}\right) = 70\,040~\text{m/s} \times \sin\left(\frac{90°}{2}\right) \approx 49\,526~\text{m/s} $$
So if your satellite-carrying spacecraft is already in Earth's orbit (which is not the same thing as an orbit around the Earth, but rather, co-orbiting the Sun with Earth), you need a total velocity change (delta-v) budget of about 54,600 m/s to enter a solar polar orbit at Venus' distance from the Sun, and that's after you apply almost 8 km/s plus drag losses to get to low Earth orbit. While there are almost certainly tricks you can use to cut down on how much of this you need to apply under power (with rocket engines running), that remains a massive undertaking. I wouldn't be the least bit surprised if you'd be looking at something similar to the Saturn C-8, which was about the same height but much bulkier than the Saturn V which sent Apollo towards the Moon.
Compare also Is it possible to communicate in space while the sun is between parties? on the Space Exploration SE.
