1. Find all positive integers $n$ that make $\frac{n^{2}}{200 n-999}$ a positive integer.
1. Let $\frac{n^{2}}{200 n-999}=k, k$ be a positive integer. Then
$$
n^{2}-200 k n+999 k=0 \text {. }
$$

Assume the equation (1) has a positive integer root $n_{1}$, and the other root is $n_{2}$. By Vieta's formulas, we have
$$
\left\{\begin{array}{l}
n_{1}+n_{2}=200 k, \\
n_{1} n_{2}=999 k .
\end{array}\right.
$$

Therefore, $n_{2}$ is also a positive integer, and $n_{1} 、 n_{2}$ both satisfy the given conditions. Without loss of generality, assume $n_{1} \geqslant n_{2}$.
From (2), we get $n_{1} \geqslant 100 \mathrm{k}$.
From (3), we get $n_{2}=\frac{999 k}{n_{1}} \leqslant \frac{999 k}{100 k}$.
Thus, $n_{2} \leqslant 9$.
Upon inspection, only $n_{2}=5$ meets the conditions, at which point, $k=25, n_{1}=4995$.
Therefore, the required $n$ is 5, 4995.