How are phased arrays compatible with the photon concept? Not a physicist (although I know how to develop and simulate phase arrays).
Phased arrays are e.g. emitters sending waves with individual elements with a different phase so that they act constructively or destructively in space and/or time and may form beams or other radiation patterns.
Clearly this is a wave phenomenon and is easily described and computed (simulated) as such.
Also clearly, is that the resulting pattern is a superposition of multiple signals from different origins that can cancel each other.
As long as there is a synchronized clock, even physically disconnected after initially being set in sync, it will work as long as the clocks remain in sync.
What I do not understand is how this phenomenon can be represented with the photon picture.
To the best of my knowledge, photons do not interact with each other.
So what is going on at the reception site that makes it seeing these interferences?
If photons were of a very higher frequency than the sources, I could imagine them as point like particles, varying in density and then there could be a count of those populations whatever their sources.
But in phase arrays, the photons frequency range is the one of the waves that are interfering with each other.
Also why the photon concept is rarely mentioned in talks about the radio part of the EM spectrum? Looks suspicious to me.
 A: 
why the photon concept is rarely mentioned in talks about the radio part of the EM spectrum?

A quick calculation shows that the photon concept is irrelevant for typical RF applications. Suppose that we have a detector that can detect some minimum power, $P$, operating at a frequency, $\nu$. The minimum number of photons it can detect is then $$N_\text{min}=\frac{P}{h\nu^2}$$ For an ultra-sensitive detector that can detect $1\text{ pW}$ at $100\text{ MHz}$ that works out to $N_\text{min}=150000$ photons as a minimum detectable threshold for a very sensitive RF detector. At those large numbers you can reliably replace any quantum mechanical operators with their expected value and use the classical operators instead.

So what is going on at the reception site that makes it seeing these interferences?

Continuing with the above numbers. Suppose that we have a phased array detector with 150 elements, and suppose that we are trying to detect a minimum-power signal. Then, even for this minimum detectable signal we expect 1000 photons per element in the array. That is already enough to reasonably use a classical approximation, but you could use a full QM approach if desired.
Photons are not localized, but have a wave function that spreads out over space. The photon wave function is governed by Maxwell’s equations, and is basically $\Psi=E+i B$. The square of this wave function gives the probability density of a photon interacting with each element of the detector. Since each detected photon gives a voltage in the element, this simply produces a signal in each element approximately as we would expect classically.
Although the phased array detector expects an average of 1000 photons per element, the actual amount will be highly non-random. Some elements may have 0 and some may have 2000, for example. A phase sensitive detector expects that, and further knows that if a specific element receives 2000 then a specific different element will receive 0. The detected signals are then amplified so that they are fully in the classical regime and then added in a phase sensitive manner. This means that if the amplified element signals match the expected pattern then the phase sensitive detector registers a signal, and if the amplified element signals do not match the expected pattern then no signal is registered. Because this involves for example hundreds of elements each detecting a thousand photons on average it is well within the classical regime. Photons bring no benefit to the analysis, even though they are there.
Edit: As mentioned above, the square of the wave function $\Psi$ gives the probability density of detecting a photon at that location. $\Psi$ behaves linearly, loosely meaning that if you have two valid wavefunctions you can add them together to get a third wavefunction which is also valid. So if you have a wavefunction for one photon and a wavefunction for another photon then the wavefunction of the system of both photons is their sum.
The probability of detecting a photon at a location is then given by the square of the sum of the wavefunctions. In locations where the two wavefunctions are out of phase they interfere and lead to a reduced magnitude. This therefore reduces the probability of detecting a photon at that location. So photon counts and interference are two ways of saying the same thing. In regions where the wavefunctions are in phase they add constructively and increase the photon count and in regions where they are out of phase they add destructively and decrease the photon count.
A: To enlarge just a bit upon Dale's most excellent answer:
There is nothing here to be suspicious about, for the following reasons. The photon model and the wave model for electromagnetic propagation represent two different ways of dealing with EM phenomena in two different wavelength regimes.
For radio waves (RF regime) the wavelengths are macroscopic (in the range of ~1000 meters to ~one centimeter) and the "classical" wave description allows straightforward solutions to be calculated for the design of antennas and so forth. Photon models furnish no practical advantage for solving problems in the RF regime.
Once your wavelengths are really small and you get out of the RF regime into light instead, the system begins exhibiting both wavelike and photon-like characteristics, and you can switch back and forth between both models depending on the particulars of your system.
As you then move to ever-shorter wavelengths, the wavelike character of the radiation becomes less readily apparent (but still present if you know how to look for it) and the photon model becomes easier to work with.
A: I have to go a little further with my answer. As you know, to generate a radio wave, you basically need an AC voltage source and a metal rod. The AC source accelerates the surface electrons on the rod back and forth and the accelerated electrons cannot help but emit photons.
With a high voltage and a strong enough source, it becomes dangerous near the rod, as the emission can be in the range of X-rays. Best seen in a radar. Let me illustrate this idea with a radar. The radar emits modulated EM radiation with frequencies from MHz to GHz. But they emit at the same time X-rays, among other things, which are in the range of 30 petahertz to 30 exahertz. What is going on?
It is simply that the changing intensity of the radiation (the radio wave) is made by the AC source, but that does not mean that the electrons emit only one photon per half-wave of the voltage source. In short, the frequency of the source is not the frequency of the photons.
So your approach

If photons were of a very higher frequency than the sources, I could imagine them as point like particles, varying in density and then there could be a count of those populations whatever their sources.

is absolutely right.
