2. For any natural numbers $n, m$ can divide $1992^{n}-999 n-1$. Then the maximum value of $m$ is ( ).
(A) 9
(B) 27
(C) 37
(D) 999
2. (D).
$\because 1999^{n}-999 n-1 \equiv 1^{n}-0-1 \equiv 0(\bmod 999)$,
$\therefore 999$ can divide $1999^{n}-999 n-1$.
$$
\therefore m_{\max } \geqslant 999 \text {. }
$$

When $n=1$, $1999^{n}-999 n-1=999$.
Then $m_{\text {max }} \leqslant 999$. Therefore, $m_{\text {max }}=999$.