The nuclei are bound and some of them are stable because of the interplay between the repulsive force of the charged protons and the residual strong force of protons and neutrons. Adding a proton increases the repulsive forces in the nucleus, adding a neutron reduces it because the neutron is not charged and it contributes only to the strong nuclear force, and it all depends on the number of protons and neutrons and the interplay of the two forces plus the Pauli exclusion principle.
The neutron when free is unstable, in contrast to the proton. It decays through the weak interaction to a proton and electron and the electron antineutrino.
The quantum mechanical nuclear shell model is successful in describing the nuclear energy levels and more or less the periodic table of elements.
I am copying from the answer here because the energy balances are explained:
Spontaneous processes such as neutron decay require that the final state is lower in energy than the initial state. In (stable) nuclei, this is not the case, because the energy you gain from the neutron decay is lower than the energy it costs you to have an additional proton in the core.
For neutron decay in the nuclei to be energetically favorable, the energy gained by the decay must be larger than the energy cost of adding that proton.
This explains why some isotopes are stable and some decay.
If neutrons keep being added as you suggest, on a stable nucleus, in order to create a heavier isotope, the probability of a neutron decaying, (nucleus beta decay) gets larger and larger, because the more neutrons, the binding energy becomes smaller and smaller compared with the energy that can be released in the decay of the neutron, and the energy needed for the proton of the decay with the extra charge.