@soothfast *said*

What you read may be right. I used to know that isolated neutrons have an average lifespan of 15 minutes and then undergo beta decay. A fist-sized chunk of neutronium would have a decent gravitational field, but not one strong enough to keep this decay from occurring irreversibly. The gravity would hold the assemblage of neutrons in a spherical shape for a very brief spe ...[text shortened]... en.wikipedia.org/wiki/Beta_decay

Perhaps DeepThought could provide deeper thoughts on the matter.

The binding energy of a nucleus is estimated by the Liquid Drop model [1]. There's a volume term, a surface area term, a coulomb term which is zero for neutronium, an asymmetry term, which is just a constant for neutronium, and a pairing term, which is of the order of 1 MeV and we can just ignore for a large enough object. This gives us:

B(A) = a_v A - a_s A^(2/3) - a_a

a_v ~ 15.8 MeV

a_s ~ 18.3 MeV

a_a ~ 23.2 MeV

Solving the cubic equation gives B(A) = 0 for A ~ 2.035 which is in accordance with observation. A quasi-bound state of two neutrons has been observed (to my surprise). Note that the pairing term will be important here.

The binding energy per nucleon is then:

B(A)/A = a_v - a_s A^(-1/3) - a_a/A

This is positive for bound states and negative for unbound states, the total energy of the nucleus is:

E = mₙc² N + mₚ c² Z - B(A, Z)

where mₙ and mₚ are the masses of the neutron and proton respectively. So the stability of neutronium depends on whether gravity can override the weak force and prevent beta decay. For small nucleii it can't, as we know. Nuclear fission is driven by the coulomb term, so there should be a threshold value where the gravitational binding energy overrides the potential energy release from beta decay. I'll have a think about how to estimate that.

[1] https://en.wikipedia.org/wiki/Semi-empirical_mass_formula