 Solution

Taking the sum once over each unordered pair, write the All-to-all Heisenberg model asFor even $N$, the minimum total spin is $S=0$, henceFor odd $N$, $S=1/2$ and $E_0=3/2-3N/2$. If the paper's ${\sum }_{ij}$ counts ordered pairs or includes $i=j$, the corresponding harmless factors and additive constant change, but the minimizing total-spin sector is the same.

Now partition an even number of spins into disjoint pairs and put every pair in a spin-one-half singlet state. Each pair has total spin zero, so their tensor product also has $S=0$ and is itself a ground state. Its internal pair contributes $-3$, while correlations between different singlets vanish, giving $-3N/2$ in total. ThereforeThe large degeneracy is special to equal all-to-all coupling: the Hamiltonian sees only total spin and cannot distinguish different singlet coverings.