Neutron capture drives iron-peak nuclei away from the valley of stability, so the nucleus that is produced has less binding energy per nucleon and is not as stable. In the slow neutron capture process (s-process), one or more neutron captures are usually followed by a beta decay, which moves the nucleus back towards the valley of stability, but now with an extra proton. In the rapid, high neutron flux r-process, then multiple neutron captures can take place before decay back towards the stability line.
The reason this works can be seen with an example. Consider this ($n,\gamma$) reaction.
$$ ^{56}_{26}{\rm Fe} + n \rightarrow ^{57}_{26}{\rm Fe} + \gamma$$
Using this semi-empirical mass formula calculator, I find that $^{56}_{26}{\rm Fe}$ has a binding energy per nucleon of 8.762 MeV and a total binding energy of 490.68 MeV. On the other hand $^{57}_{26}{\rm Fe}$ has a binding energy per nucleon of 8.728 MeV and a total binding energy of 497.48 MeV.
Thus although the binding energy per nucleon is smaller in the product nucleus, the total binding energy is larger. If we consider only the rest mass of the neutron then the total mass-energy on the LHS of the reaction is 53030.91 MeV, whilst on the RHS it is 53024.26 MeV. Thus there is an excess of available mass-energy that is carried off by the gamma photons and/or in the respective kinetic energies of the particles.
A side issue is that the statement that a particular A/Z ratio is the most stable depends on the density and environment of the material. For example in the collapsing core of a pre-supernova star, electron degeneracy blocks beta decay and pushes the equilibrium to more neutron-rich nuclei and this aids neutron capture.
Finally, the reason why neutron captures can occur at a significant rate (in the presence of a high enough neutron flux) but that fusion reactions do not, is that (neutral) neutrons do not suffer the coulomb repulsion that prevents the initiation of alpha capture and other fusion reactions.