China Girls' Mathematical Olympiad

It is easy to see that $n$ is an even number. Let $n=2m$, then from $40m+2\mid 2003\times 2m+2002$ we can get $20m+1\mid 2003m+1001$. But $2003m+1001=100(20m+1)+3m+901$, so $20m+1\mid 3m+901$. And when $\frac{3m+901}{20m+1}=1,2,3,4$, $m$ is not a positive integer. Therefore, $$\begin{array}{rl}\frac{3m+901}{20m+1}& \ge 5,\\ m& \le \frac{896}{97}<10.\end{array}$$ Hence $m\le 9$. But after checking one by one for $m=1,2,3,\dots ,9$, we know that $20m+1\nmid 3m+901$ for $m=1,2,3,\dots ,9$. Therefore, there is no positive integer $n$ satisfying the condition.