= Solution
For $g\in\mathcal H_N$, write
$$
\widehat g(\alpha)=\sum_{n\leq N}g(n)e(\alpha n).
$$
The <orthogonality of complex exponentials> converts the <linear configuration count> into
$$
T(g,g,g,g)=\int_0^1\widehat g(\alpha)\widehat g(2\alpha)\widehat g(3\alpha)\widehat g(-4\alpha)\,d\alpha.
$$
Expand the difference between the products for $g_1$ and $g_2$ by changing one factor at a time. A typical term is
$$
\int_0^1\bigl(\widehat g_1(\alpha)-\widehat g_2(\alpha)\bigr)\widehat h_2(2\alpha)\widehat h_3(3\alpha)\widehat h_4(-4\alpha)\,d\alpha,
$$
where each $h_j$ is either $g_1$ or $g_2$. The assumed <uniform norm> bound controls the first factor by $\varepsilon N$. The substitution $\alpha\mapsto k\alpha$ preserves an integral over the <circle group>, so <Hölder's inequality> and the three supplied $L^3$ bounds give
$$
\int_0^1\prod_{j=2}^4|\widehat h_j(k_j\alpha)|\,d\alpha
\leq\prod_{j=2}^4\left(\int_0^1|\widehat h_j(\alpha)|^3\,d\alpha\right)^{1/3}
\ll N^2.
$$
Each of the four terms is therefore $O(\varepsilon N^3)$, and hence
$$
\boxed{|T(g_1,g_1,g_1,g_1)-T(g_2,g_2,g_2,g_2)|=O(\varepsilon N^3).}
$$
This is <Fourier stability of a linear configuration count>.
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