Find all integers $n$ such that $20n+2$ divides $2003n+2002$.

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Solution: Since $20 n+2$ is even and can divide $2003 n +2002$, therefore, $2003 n+2002$ is even. Thus, $n$ is even.
Let $n=2 m$ (where $m$ is an integer), then
$$
\begin{array}{l}
\frac{2003 n+2002}{20 n+2} \in \mathbf{Z} \\
\Leftrightarrow \frac{2003 m+1001}{20 m+1} \in \mathbf{Z} \\
\Leftrightarrow \frac{100(20 m+1)+3 m+901}{20 m+1} \in \mathbf{Z} \\
\Leftrightarrow \frac{3 m+901}{20 m+1} \in \mathbf{Z} \\
\Leftrightarrow \frac{20(3 m+901)}{20 m+1} \in \mathbf{Z}(20 \text { and } 20 m+1 \text { are coprime }) \\
\Leftrightarrow \frac{3(20 m+1)+18017}{20 m+1} \in \mathbf{Z} \Leftrightarrow \frac{18017}{20 m+1} \in \mathbf{Z} \\
\Leftrightarrow \frac{43 \times 419}{20 m+1} \in \mathbf{Z}(43 \text { and } 419 \text { are both prime numbers }) \\
\Leftrightarrow 20 m+1= \pm 1, \pm 43, \pm 419, \pm 18017 \\
\Leftrightarrow m=0,-21 \Leftrightarrow n=2 m=0 \text { or }-42 .
\end{array}
$$

Therefore, the integers $n$ sought are 0 or -42.
(Wu Weizhao, School of Mathematics and Information Science, Guangzhou University, 510006)