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,
$$\begin{aligned} \frac{3m+901}{20m+1} &\ge 5, \\ m &\le \frac{896}{97} < 10. \end{aligned}$$
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.