= Solution
Add the one-particle <energy> term $H_0=\epsilon_1n_1+\epsilon_2n_2$. It is diagonal in the same <Fock state> basis as the <contact interaction between two ring modes>. For $N=2$, the ordered basis $(|2,0\rangle,|1,1\rangle,|0,2\rangle)$ gives
$$
\boxed{H=\begin{pmatrix}
2\epsilon_1+g/L&0&0\\
0&\epsilon_1+\epsilon_2+2g/L&0\\
0&0&2\epsilon_2+g/L
\end{pmatrix}.}
$$
For $g>0$, the middle <eigenvalue> exceeds the mean of the two endpoint <eigenvalues> by $g/L$, so it cannot be the smallest. The <ground state> is therefore \b[$|2,0\rangle$ if $\epsilon_1<\epsilon_2$, and $|0,2\rangle$ if $\epsilon_2<\epsilon_1$], with
$$
\boxed{E_{\min}=2\min(\epsilon_1,\epsilon_2)+g/L.}
$$
If $\epsilon_1=\epsilon_2$, every normalized <linear combination> of the two endpoint <Fock states> is a <ground state>. The same endpoint argument works for general fixed $N$, since the interaction contribution is a concave quadratic function of the mode <occupation number>.
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