= Solution
Put $\eta=\operatorname{Im}z>0$. For a real <eigenvalue> $\lambda$,
$$
\operatorname{Im}\frac1{\lambda-z}=\frac{\eta}{(\lambda-\operatorname{Re}z)^2+\eta^2}>0.
$$
The <spectral theorem for real symmetric matrices> therefore gives $\operatorname{Im}g>0$ and $\operatorname{Im}q_i\geq0$, since $q_i$ is a sum of $(\lambda-z)^{-1}$ with nonnegative coefficients given by squared eigenvector coordinates of $x_i$. This is the <positive imaginary part of a Stieltjes matrix resolvent>. Hence
$$
|z+g|\geq\eta,\qquad |z+q_i|\geq\eta.
$$
Apply the <triangle inequality> to the identity above, and add and subtract the minor trace:
$$
\begin{aligned}|\varepsilon_N(z)|&\leq\frac1{\eta^2N}\sum_{i=1}^N|q_i-g|\\&\leq\frac1{\eta^2N}\sum_{i=1}^N|q_i-g^{(i)}|+\frac1{\eta^2N}\sum_{i=1}^N|g^{(i)}-g|.
\end{aligned}
$$
The allowed <principal minor resolvent trace bound>, with the printed $N$ normalization, bounds the second average by $c/(\eta N)$. Thus the explicit conclusion is
$$
\boxed{|\varepsilon_N(z)|\leq\frac1{(\operatorname{Im}z)^2N}\sum_{i=1}^N\left|x_i^T(X_N^{(i)}-zI)^{-1}x_i-g_N^{(i)}(z)\right|+\frac{c}{(\operatorname{Im}z)^3N}.}
$$
In particular, the final term is $O((\operatorname{Im}z)^{-3}N^{-1})$ with a constant independent of $N$. The upper-half-plane restriction gives the displayed positive denominators; in the lower half-plane the analogous bound uses $|\operatorname{Im}z|$.
The chosen centering also has a probabilistic meaning. The column vector $x_i$ is independent of the minor, has mean zero and coordinate <variance> $1/N$, so its <conditional expectation> satisfies
$$
\mathbb E[q_i\mid X_N^{(i)}]=\frac1N\operatorname{Tr}G^{(i)}=g_N^{(i)}.
$$
This explains why the bound isolates fluctuations of a <quadratic form> around the minor trace. Mean centering alone is not a concentration estimate; no unrequested limiting assertion is being assumed.
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