The closest you will ever come in nature to pure neutron matter is the nuclear matter in neutron stars, but it is not pure either, being suffused with protons and electrons. Neutrons are 1.4 MeV heavier than protons and tend to undergo beta decay to $pe\bar{\nu }$, unless stabilized by the exclusion principle and an abundance of protons, electrons, or antineutrinos in the vicinity. Were it not for the Coulomb repulsion between protons, ordinary nuclear matter would have more protons than heavier neutrons!
Let us concentrate on nuclear matter. Recall that in a non-interacting Fermi gas, all momentum states below the spherical Fermi surface of radius ${{p}_{F}}$ are filled. The condition for overall electrical neutrality is $p_{p}^{3}=p_{e}^{3}$, based on the volume inside the Fermi sphere. At zero temperature, we may identify the chemical potential with the corresponding energy, $\mu ={{E}_{F}}=\sqrt{p_{F}^{2}+{{m}^{2}}}$. The condition for chemical equilibrium is ${{\mu }_{n}}={{\mu }_{p}}+{{\mu }_{e}}+{{\mu }_{{\bar{\nu }}}}$. Interactions roil the Fermi surface but don’t actually change the equilibrium significantly. Since nuclear forces are independent of charge, they do not distort the difference ${{\mu }_{n}}-{{\mu }_{p}}$.
When a neutron star forms, most protons undergo inverse beta decay, and it seems safe to assume that the neutrinos escape, hence ${{\mu }_{{\bar{\nu }}}}=-{{\mu }_{\nu }}=0$. We now have enough conditions to determine the proportions of neutrons, protons, and electrons at any given density.
The chemical potential of neutrinos is more relevant cosmologically. If God had assigned the creation of the universe to his most incompetent archangel-trainee, who then poured too many antineutrinos into the mix, you might very well have nothing but pure neutron matter, and you wouldn’t be here to ask about it. (Hmmm ... neutrons instead of neurons ... even worse than having rocks in one’s head.)
At densities well above what is expected in neutron stars, nucleons would overlap, so it would be more accurate to think in terms of quark matter, comprising u and d quarks in three colors. The conditions become ${{\mu }_{d}}={{\mu }_{u}}+{{\mu }_{e}}+{{\mu }_{{\bar{\nu }}}}$ and $-p_{e}^{3}+\tfrac{2}{3}(3)p_{u}^{3}-\tfrac{1}{3}(3)p_{d}^{3}=0$.
