Why can the neutron not be measured in this fixed-target experiment? Consider a fixed-target-experiment, where negatively charged pions are shot at protons (the latter being at rest). The kinetic energy of the pions shall be known. One possible reaction is
$$
\pi^- + p \to n + \pi^+ + \pi^-
$$
creating a neutron. The neutron apparently remains undetected by measuring devices, and only one of the pions can be measured.
I have two conceptual questions:
(1) Why is this? (I mean, both pions carry charge...)
(2) If the neutron cannot be measured, then how can we be sure that it is there?
 A: Exclusive & semi-inclusive reactions are commonly as written target(beam, detected)undetected:
$$p(\pi^-, \pi^+)n\pi^-$$
To understand the detection of the final state, we really need more information. While there are detectors that can handle multiple particle final states (https://www.jlab.org/physics/hall-b/clas), this experiment wasn't done there.
More commonly, fixed target experiments put a beam on target and look at reaction products with a spectrometer (e.g., SLAC's 8 GeV spectrometer: https://www.slac.stanford.edu/pubs/slacpubs/5750/slac-pub-5753.pdf). This grabs a small bite of both angular space and momentum space. It uses (a) dipole magnet(s) to select a central momentum, and quadruple magnets to focus particle tracks so that momentum and lab-angle (regardless of out-of-plane angle) can be determined.
The spectrometer can only look a one charged final state, but that can be reversed by reversing the current. Since the beam is $\pi^-$, the $\pi^+$ final state is much cleaner.
The way to separate the reaction from:
$$p(\pi^-, \pi^+)X$$
is the invariant mass of $X$. The initial state's 4-momentum is:
$$p_i^{\mu} = p_p^{\mu} + p_p^{\pi^-} $$
$$p_i^{\mu} = (M_p, 0,0,0) + (E_{\pi^-}, 0,0,p_{\pi^-}) $$
while the final state is:
$$p_f^{\mu} = p^{\mu}_{\pi^+} + p_X^{\mu}   $$
Of course, $p_i^{\mu}= p_f^{\mu}$ so that:
$$p_X^{\mu} = p_i^{\mu} - p^{\mu}_{\pi^+} $$
$$p_X^{\mu} = (M_p+E_{\pi^-}, 0,0, p_{\pi^-}) - (E_{\pi^+}, p_{\pi^+}\sin{\theta},0, p_{\pi^+}\cos{\theta}) $$
If you work that all out at a fixed angle, there should be a momentum region that ensures $X$ is a nucleon plus a pion, and since it has charge -1, it has to be a neutron and $\pi^-$. (Another reason to detect the positive pion, as a negative pion in the final state does NOT distinguish between an undetected $n\pi^+$ and $p\pi^-$). That also rules out a lot of other final states, so by detecting  the $\pi^+$, there is a kinematic region where $X=n\pi^-$ is the only viable option.
