Why do neutrons repel each other?

Neutrons (and protons) being spin 1/2 fermions, must fit antisymmetric wavefunctions. This "wavefunction" doesn't always involve waves, though. For nucleons - the generic term for neutron or proton - this wavefunction for the pair is a produce of (1) a spatial part, (2) a spin part, and (3) an isospin part.

The isospin part is a clever way to describe charge possibilities of otherwise identical particles. We regard neutrons and protons as being in a sense identical. Just as a spin 1/2 particle can be "up" or "down" along some chosen axis, so is an isospin 1/2 particle can be "up" or "down" along an abstract mathematical axis - it's exactly the same SU(2) math as spin - but it plays out in physical reality as charge. For nucleons it's not +1/2 and -1/2 charge but with an offset, so we have +1 (proton) and 0 (neutron). This idea is from Heisenberg in 1932.

Now, how can the overall wavefunction of a pair of particles be antisymmetric? There are three factors - right away we can imagine three possibilites: any one factor being antisymmetric with the other two symmetric. We could also have all three antisymmetric at the same time.

An antisymmetric spatial wavefunction would have a node, like an atomic p orbital, like the electric potential around a dipole antenna. This is a higher energy state than a simple spherical blog, a Gaussian. Given the range of internuclear forces, this nodal antisymmetric wavefunction has more energy than if the two nucleons just stayed apart. This is a matter of radial or angular kinetic energy has to be either "zero" or some quantized value that exceeds "escape velocity" So forget that part of the system wavefunction being antisymmetric.

BTW, we don't have separate spatial wavefunctions for the two nucleons - whatever one does, the partner does the exact opposite, like a two-body celestial mechanics problem. They orbit a common barycenter.

The spin part could be antisymmetric. This is a bit tricky. If particle #1 is up and #2 is down, we can write "UD". There is also "DU". We form the spin part of the wavefunction for the pair as UD-DU. We could instead choose UD+DU but note this is symmetric. So are UU and DD. Just how UD-DU differs from UD+DU may mystify beginners in quantum mechanics, but it's important, and it's how physical matter works whether we humans like it or not. (You might also see where the 'u' and 'd' quarks got their names. The quark idea came along years after isospin.)

Neither D nor U is really a wave or a function; they're at most just rows and columns in matrices if you must represent them in familiar math. Otherwise quantum physicists just deal with these symbolically. Still the jargon is "wavefunction" - we silly humans and our primitive scientific language!

The same math applies for isospin. But the physics differs. We've concluded that the spatial part of the system wavefunction must be symmetric, so it's up to either the spin part or the isospin part to be antisymmetric. But not both! If spins are symmetric - they're parallel. This is experimentally the case - the Deuteron (obtain by distilling "heavy water" from water) - so we deduce that the isospin part is antisymmetric. That is, we must have one isospin "up" and one isospin "down" - neutron and a proton, not two neutrons or two protons.

Just why must the spins of the two nucleons be parallel? The strong force holding them together - the exchange of pions, kaons and other mesons - works better in that case. To explain that takes deeper analysis than I can go into here. When the spins are antiparallel, there's not enough force to keep the nucleons together.

This would be the case though, were to try pushing two neutrons together. They'd be both isospin "up" therefore symmetric isospin part of the wavefunction, therefore requiring an antisymmetric spin part, which leads to the pions and their buddies not getting as good a grip on the neutrons, which drift off going their separate ways.

Neutrons have spin 1/2 and therefore obey the pauli exclusion principle, meaning two neutrons cannot occupy the same space at the same time. When two neutrons' wavefunctions overlap, they feel a strong repulsive force. See http://en.wikipedia.org/wiki/Exchange_interaction .

I think there are two parts to this answer. The first is to do with an ensemble of neutrons in a dense fermion gas and the second is to do with the strong nuclear force between two neutrons (in a many-body nucleon system).

Neutrons in a dense gas will be degenerate. That is to say that the Pauli exclusion principle prevents more than two neutrons (spin up and spin down) occupying the same momentum eigenstate. This means that even at zero temperature, neutrons in the gas can have very large momenta up to the Fermi momentum. In fact in the centre of a typical neutron star, the neutron kinetic energy is becoming comparable with the neutron rest mass-energy even when it is "cold". The momentum of the neutrons leads to a degeneracy pressure that is able to (partially) support the weight of a neutron star.

However, pure degeneracy pressure actually ignores the interactions between the particles - it assumes they are ideal, non-interacting and point-like. From that point of view the neutrons can get very close together (and indeed @Marek they do "nestle" up to each other at densities approaching $10^{18}$ kg/m$^3$, the PEP does not forbid that because it is phase space [momentum x volume] that they have to be separated in, not just physical space).

But interactions do become important at these high densities. In a dense, many body nucleon gas (there are some protons in a neutron star too) the strong nuclear force is attractive at "long range" (where this means $>10^{-15}$m!) but becomes repulsive at shorter ranges. This latter hardens the equation of state (pressure increases more quickly with increasing density), allowing neutron stars to have much higher masses than would be allowed for ideal neutron degeneracy pressure alone. Ideal NDP would only allow neutron stars up 0.7Msun before collapse to a black hole, but with interactions they could become as high as 3Msun (a very uncertain number).

This doesn't really explain why n-n forces become repulsive at small separations (see DarenW's answer for that, though the situation becomes more complex and is less well understood in high density nucleon gases), but does I think clarify some confusion on this thread between the role of degeneracy pressure and the role of the strong nuclear force.

Edit: It is quite likely that at extremely high densities that the neutrons will undergo a phase transition. Firstly, their Fermi energies become large enough that the creation of other particles (e.g. hyperons or mesons) becomes possible. Secondly, the quarks in the neutrons may achieve asymptotic freedom and the neutrons "dissolve" into a quark-gluon plasma. Neither possibility is theoretically secure or has unambiguous observational evidence and it is unclear whether a stable neutron star can have a core featuring these phases or whether collapse to a black hole will already be in progress.