= Solution
For distinct positive integers $p,q$, use normalized box modes $\varphi_1=\sqrt{2/L}\sin(p\pi x/L)$ and $\varphi_2=\sqrt{2/L}\sin(q\pi x/L)$. Their <contact interaction> overlaps obey
$$
I_{1111}=I_{2222}=\frac{3}{2L},\qquad I_{1122}=\frac1L.
$$
Unlike traveling ring modes, these standing modes also allow pair conversion. For the first two box modes, or any distinct pair without a third-harmonic resonance, the <contact interaction between two box modes> is
$$
\boxed{H_{\rm int}=\frac{3g}{4L}\left[n_1(n_1-1)+n_2(n_2-1)\right]
+\frac{2g}{L}n_1n_2+\frac{g}{2L}\left(a_1^{\dagger2}a_2^2+a_2^{\dagger2}a_1^2\right).}
$$
The last two terms change an individual mode <occupation number> by two while conserving total <particle number>. Thus the <Hamiltonian operator> is not diagonal in the <Fock state> basis, and minimizing diagonal entries alone does not find its <ground state>. For example, at $N=2$ with no dispersion, $\left(|2,0\rangle-|0,2\rangle\right)/\sqrt2$ has the lowest <energy> $g/(2L)$, below either unmixed endpoint's $3g/(2L)$.
The wording permits modes beyond the first two. For completeness, the identity $\sin^3u=(3\sin u-\sin3u)/4$ gives
$$
I_{1112}=-\frac{\delta_{q,3p}}{2L},\qquad I_{1222}=-\frac{\delta_{p,3q}}{2L}.
$$
When $q=3p$, add $-g(a_1^{\dagger2}a_1a_2+a_1^\dagger a_2^\dagger a_1^2)/(2L)$; when $p=3q$, add $-g(a_1^\dagger a_2^\dagger a_2^2+a_2^{\dagger2}a_1a_2)/(2L)$. These extra terms change an individual <occupation number> by one and require diagonalizing the corresponding fixed-$N$ matrix as well.
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