1. Given $1998^{n} \mid 1999$!, then the maximum value of the positive integer $n$ is
$=1.55$.
$$
\because 1998=2 \times 3^{3} \times 37,3^{3}<37 .
$$
$\therefore 1999$ ! contains the factor of the form $37^{m}$, where the maximum value of $m$ is the maximum value of $n$.
$$
\therefore n \leqslant \sum_{i=1}^{\infty}\left[\frac{1999}{37^{i}}\right]=55 \text {. }
$$

Therefore, the maximum value of $n$ is 55.