1. If the set $S=\{n \mid n$ is an integer, and $22 n+2$ divides $2003 n+2004\}$, then $S$ is ( ).
(A) empty set $\varnothing$
(B) singleton set
(C) set with two elements
(D) infinite set
-、1.(C).
Obviously, $2 \mid n$. Let $n=2k$, then $(22k+1) \mid (2003k+1002)=91(22k+1)+k+911$.
Thus, $(22k+1) \mid (k+911)$, which is equivalent to $(22k+1) \mid (22k+911 \times 22)=(22k+1)+20041$, that is,
$$
(22k+1) \mid 20041=7^2 \times 409 \text{, }
$$

where 409 is a prime number.
Therefore, $22k+1$ can take $\pm 1, \pm 20041, \pm 49, \pm 409$, $\pm 7 \times 409$. After inspection, $k$ can only take 0 and -911. Hence, $n=0$ and -1822. Therefore, $S$ is a two-element set.