The strong force has an approximate symmetry called isospin, which is a rotation in an abstract space spanned by the neutron and proton. This symmetry is broken by the Coulomb force, which only acts on protons, and by the slight mass difference between up quarks and down quarks, which makes the neutron slightly heavier than the proton.
Consider possible bound states of neutrons and protons. These are fermions, so the wave function has to be anti-symmetric. Neutrons and protons are heavy, so we can analyze the situation using non-relativistic quantum mechanics. If a bound state exists, the ground state is in an s-wave (symmetric) because the centrifugal barrier is purely repulsive. This means that the spin-isospin wave function has to be anti-symmetric. This leaves two possibilities:
$$
(I=0,S=1) \;\;or\;\; (I=1,S=0)
$$
where $I$ and $S$ are the total isopsin and spin of the state. Note that $I,S=0$ is anti-symmetric in isospin/spin, and $I,S=1$ is symmetric. The $I=0$ state corresponds to
$$
|I=0\rangle = \frac{1}{\sqrt{2}}\left(|np\rangle -|pn\rangle\right)
$$
and $I=1$ has three isospin components
$$
|I=1,I_3=+1,0,-1\rangle = \left\{|pp\rangle ,\frac{1}{\sqrt{2}}\left(|np\rangle +|pn\rangle\right),|nn\rangle\right\}
$$
The potential in the $I=0,1$ channels is determined by complicated interactions between quarks and gluons.
Empirically (or from numerical calculations in lattice QCD) we know that both the $I=0$ potential and the $I=1$ potential are attractive, but the $I=0$ interaction is slightly more attractive.
Note that in non-relativistic QM (in three dimensions), an attractive interaction is not sufficient to form a bound state. The potential has to have a minimum strength. Nuclear physics is fine-tuned, the empirical NN potentials are close to the threshold where bound states appear.
There is an $I=0$ bound state, called the deuteron (binding energy 2.2 MeV), but the di-neutron (and its isospin partners) fall just short of being bound. Due to the Coulomb force, there is a little extra repulsion in $pp$, but the $nn$ channel is almost bound.
This can be quantified using the scattering length. The $nn$ scattering length is about 20 fm, and the corresponding energy scale is
$$
B=\frac{1}{2ma^2}\simeq 50 \, keV
$$
so the di-neutron comes within 50 keV of being bound. This also means that if one could change the quark masses slightly, there would presumably be a bound di-neutron (this can be checked in lattice QCD).
It is possible in principle to have a three neutron bound state even if there is no two neutron bound state (states like that are called Borromean), but this is disfavored by the Pauli principle (not all three neutrons can be in an s-wave). The same applies to 4n bound states (there is a recent claim of a 4n resonance http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.116.052501).
In practice $^3H=pnn$ and $^3He=ppn$ exist, but not $3n$.
Note that neutron stars, clusters of $10^{57}$ neutrons, do exist. Of course, these objects are gravitationally bound.
Best Answer
When a nucleus decays the reaction is characterised by the release of a fixed amount of energy called the Q-value of the reaction.
This diagram shows what happens when AM-241 decays to Np-237 with the emission of an alpha particle.
The energy levels in both nuclei are well defined and so the energies of the alpha particles are well defined.
$\rm energy_{\rm decay} =energy_{excited \,daughter} + energy_{alpha}$
So in the example shown the energies of the emitted alpha particles will be 5.48, 5.54, 5.58, 5.61 and 5.64 Mev.
The excited daughter nucleus then gets rid of the surplus energy with the emission of a gamma.
However for beta decay the quantum jumps as characterised by those shown below for alpha particles are accompanied by the emission of two particles which together carry away a fixed amount of kinetic energy.
The difference is that the sum of the kinetic energy (beta) and the kinetic energy (anti neutrino) is fixed but the energies of the emitted particles is not.
This means that you can have a range of beta energies and a corresponding range of anti neutrino energies.
In this example there would be a range of beta energies from zero (approx) to 0.31 MeV for one decay mode and from zero to 1.48 Mev for the other decay mode. The rest of the energy being taken away by the antineutrino.
$\rm energy_{\rm decay} =energy_{excited \,daughter} +energy_{\rm beta}+ energy_{\rm antineutrino}$
Again the excited daughter nucleus gets rid of the surplus energy via gamma emission
So it is the maximum energy of the beta particles which is the characteristic of the decay.