= Solution
Write the equally spaced one-particle spectrum as
$$
E_n=E_0+n\Delta.
$$
The two-particle level labelled by $N$ has energy
$$
E^{(2)}_N=2E_0+N\Delta
$$
and receives contributions from orbital pairs $(n_1,n_2)$ with $n_1+n_2=N$.
For every unequal unordered pair $n_1<n_2$, there is one symmetric and one antisymmetric spatial combination. The symmetric one combines with the six-dimensional symmetric spin space; the antisymmetric one combines with the three-dimensional antisymmetric spin space. Thus each unequal pair contributes nine states.
If $N$ is odd, all pairs are unequal and there are $(N+1)/2$ unordered pairs. Therefore
$$
\boxed{g_N=\frac{9(N+1)}2,
\qquad N\text{ odd}.}
$$
If $N$ is even, there are $N/2$ unequal pairs and the additional equal pair $n_1=n_2=N/2$. The latter has only a symmetric spatial state and therefore contributes six spin states. Hence
$$
\boxed{g_N=9\frac N2+6,
\qquad N\text{ even}.}
$$
This is the <degeneracy of two identical spin-one bosons with equally spaced orbital levels>.
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