Why are neutrons preferred for measuring cosmic flux (Forbush decreases)? I see (I asked this question, today read Wikipedia's article on 'neutron monitor', and other stuff, they're always neutrons) that neutrons are the favourite secondary cosmic ray when studying Forbush decreases, periodicities in the cosmic flux, etc. I know people detect other particles also, but I don't see those when covering that topics. Neutrons are harder to detect, so there must be a good reason for not preferring . . . I don't know, protons, or electrons?
Are they more abundantly produced, so they give better statistics? (I would say they aren't.)
The linked Wikipedia page describes neutron monitors' measurement technique, that "keeps out environmental, non-cosmic ray induced neutrons" and "amplifies the cosmic signal". Can't we do better with charged particles?
 A: As you've noted in a comment, neutrons from cosmic rays are secondary cosmic rays, produced through collisions (a type of spallation) with primary cosmic rays with atmospheric particles (typically $\text{N}_2$ or $\text{O}_2$). Neutrons have such short lifetimes ($\sim15$ minutes), and so it is impossible for them to travel cosmic distances.
When a cosmic ray - generally a proton, but sometimes an alpha particle or a more massive nucleus - hits an atom in the atmosphere, various particles can be produced, including protons, neutrons, and various mesons and leptons. Here's a table of possible products (from here):
$$\begin{array}{|c|c|c|c|c|}
\hline \text{Particle}&\text{Rest mass}&\text{Mean lifetime}&\text{Decay mode}&\text{Decay frequency}\\
\hline \text{p}&940\text{ MeV}&>10^{25}\text{ Years}&\text{N/A}&\text{N/A}\\
\hline
\hline \text{n}&940\text{ MeV}&887\text{ s}&\text{p}+e^-+\bar{\nu}_e&100\%\\
\hline \pi^{\pm}&140\text{ MeV}&26\times10^{-9}\text{ s}&\mu^{\pm}+\nu_{\mu}&99\%\\
\hline \pi^0&130\text{ MeV}&8\times10^{-17}\text{ s}&2\gamma&99\%\\
\hline K^{\pm}&500\text{ MeV}&12\times10^{-9}\text{ s}&\mu^{\pm}+\nu_{\mu},\quad\pi^{\pm}&63\%,\quad27\%\\
\hline \mu^{\pm}&110\text{ MeV}&2.2\times10^{-6}\text{ s}&e^-+\bar{\nu}_e+\nu_{\mu}&99\%\\
\hline \nu_{\mu}\text{ }^{\dagger}&<1\text{ eV}&?&\text{N/A}&\text{N/A}\\
\hline e^{\pm}&0.51\text{ MeV}&>10^{23}\text{ years}&\text{N/A}&\text{N/A}\\
\hline \nu_{e}\text{ }^{\dagger}&<1\text{ eV}&?&\text{N/A}&\text{N/A}\\
\hline
\end{array}$$
$^{\dagger}$ I've made some changes to the table based on newer experiments. Even smaller limits have been placed on the masses of neutrinos, and the known "decay" paths of neutrinos aren't really decays at all, but flavor oscillations. I've also denoted which particles are antiparticles (in some of the decay chains). However, I've kept the rounding of masses and lifetimes; the rounding is insignificant here.

Masses and energies
The problem with certain secondary cosmic ray particles is that they are produced in large numbers, with low mean energies. Let's imagine that a proton ($\text{p}$) collides with an oxygen nucleus ($N$) in a molecule of $\text{O}_2$. It could produce a reaction of the form
$$N+\text{p}\to N+\text{n}+\pi^++\pi^-+\pi^0+e^++\nu_e+\text{photons}$$
This was clearly a very energetic gamma ray!$^*$ Now, very quickly, the two positive pions ($\pi^+$) decay into anti-muons ($\mu^+$) and muon neutrinos ($\nu_{\mu}$). The anti-muons then decay further, into positrons, electron neutrinos, and muon neutrinos. The negative pion ($\pi^-$) goes through the same process, but with those particles' antiparticles. The neutral pion ($\pi^0$) decays into two photons ($\gamma$), which, along with the other photons produced in the original reaction, may have enough energy to each form electron-positron pairs. All of these final products (electrons, neutrinos, and their antiparticles) are now stable, and the neutron has a good chance of not decaying before reaching the ground.
We now have quite a lot of leptons and anti-leptons moving around. However, because of conservation of energy, each of these particles has much less energy than the original cosmic ray, meaning that they are hard to detect. As the linked page writes,

At sea level, for every 10,000 muons, there will still be roughly: 200 primaries (protons and occasional neutrons), 20 high-energy electrons (E>1GeV), and 4 pions. But there may be up to 100,000 low energy electrons created by the cascade. These particles are absorbed quickly, but if the shower is energetic enough or the shower started low enough, they may still be the most prevalent particles at sea level. However, because of their lack of penetrating power, the plastic scintillators we are using, won’t detect them anyway.

So that gets rid of a lot of possible candidates.
Eliminates (on the whole):


*

*Pions

*Muons

*Neutrinos

*Electrons


$^*$ We see something akin to positron emission here, insofar as a proton goes in and a neutron goes out (do not take the analogy too far!). Normally, this might seem odd because extra mass-energy would be needed, but there is quite a lot of energy to go around. Cosmic ray energies are measured in the $\text{GeV}$ range, and you only need a small fraction of that to make up the difference between the rest masses of a neutron and a proton! Also, not all these decay chains are going to happen in every case; they are simply example processes here.

Mean lifetime
The pions, kaons, and muons all have lifetimes on the order of or less than $\sim10^{-6}$ seconds. That means that most will quickly decay before reaching the ground - although keep in mind that these are mean lifetimes, not absolute lifetimes. Additionally, time dilation at these speeds can be incredibly significant, and means that many muons will reach the ground. However, it's still always better to have particles with longer mean lifetimes, as fewer will decay over the same timescales.
Say we have a particle of energy $E$ and rest mass $m$. The Lorentz factor, $\gamma$, is
$$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$$
and the total energy is
$$E=\gamma mc^2$$
Therefore,
$$\gamma=\frac{E}{mc^2}$$
Let's say we have a primary cosmic ray with energy on the order of $\sim10\text{ GeV}$, and a muon produced via pion decay is imbued with $10\%$ of that energy. We then find that
$$\gamma\approx10$$
The muon will undergo time dilation, with its lifetime increasing by a factor of 10 when seen by an external observer. Now, given that $\gamma\propto E$, a kaon would need to have a Lorentz factor of roughly $\sim10^2$-$10^3$ times that of the muon to have a lifetime of the same order of magnitude, and pions would need Lorentz factors something like $10^8$ times that. A look at a primary cosmic ray energy distribution shows that even incoming protons or nuclei are unlikely to have that sort of energy, before colliding with atmospheric nuclei. The upshot of this is that only muons, out of the decay-prone leptons or mesons, have a shot at reaching the ground.
Eliminates:


*

*Pions

*Kaons

*(Some) muons



Interactions with various things
If the energy criterion wasn't enough, we can be assured that neutrinos are a terrible choice. They interact with other particles primarily through the weak nuclear force, which is . . . well, weak. That's why it's so difficult to detect them at all, let alone in significant quantities.
Additionally, as was discussed in the comments, magnetic fields do funky things to charged particles. In this case, protons and electrons can be redirected by Earth's magnetic field (not the Sun's) before entering the atmosphere. This isn't really an issue if we're looking at secondary particles, but it does mean that charged particles often aren't good choices (if they originate outside the atmosphere).
Eliminates:


*

*Neutrinos

*(Some) protons

*(Some) electrons



Nucleon energies
Here's where we get to the brunt of your question: Why not protons? After all, they comprise the majority of primary cosmic rays ($\sim90\%$, I believe), and may be produced in larger quantities than neutrons in certain atmospheric collision-based reactions. The thing is, most of the primary cosmic ray protons must have high energies in order to reach the surface in large quantities. This means that at lower energies, there should be a higher neutron-to-proton ratio.
This might be unimportant, but evidence suggests that Forbush decreases may be more prominent when influencing low-energy particles, as opposed to high-energy particles - something that was known in the 1950s (see Simpson (1957) and Lockwood (1971), for an earlier review)! Newer experiments continue to show this effect (see Ifedili (2007)).
Eliminates:


*

*(Some) protons



There are a host of particles that can be produced by cosmic rays through interactions with the atmosphere, but most of them are poor choices for any of a variety of reasons. They may not have enough energy to be as easily detected in large quantities (pions, muons, electrons, and neutrinos), may have too short lifetimes (kaons and pions), may interact too weakly with matter (neutrinos), may be deflected by the Earth's magnetic field (primary cosmic rays of protons and electrons), or may have too high energies (protons). Neutrons don't fall victim to any of these issues.
I'm not saying that it's impossible to detect any of the other particles. Ambient muons may be detected by a small, portable cloud chamber that can be easily made in a classroom (pro-tip: do not shield it from said muons), for instance. It's simply that it made be better and/or easier to detect neutrons than other particles. So when I write "Eliminates", it doesn't mean that it's impossible to detect these particles; it just means that they're not anyone's top choice.
