11.5 Find the smallest positive integer that can be expressed as the sum of 2002 positive integers with equal sums of their digits, and also as the sum of 2003 positive integers with equal sums of their digits.
11.510010 .

Assume for a positive integer $n$ the following expression holds:
$$
n=a_{1}+a_{2}+\cdots+a_{2002}=b_{1}+b_{2}+\cdots+b_{2000} \text {. }
$$

Notice that the sum of the digits of $a_{1}, a_{2}, \cdots, a_{2002}$ is the same, so they have the same remainder when divided by 9. Let this remainder be $r(0 \leqslant r \leqslant 8)$. Similarly, the remainder when $b_{1}, b_{2}, \cdots, b_{2003}$ are divided by 9 is the same, denoted as $s(0 \leqslant s \leqslant 8)$.
Thus, $n-2002 r$ and $n-2003 s$ are both multiples of 9. Therefore,
$$
\begin{array}{l}
(n-2002 r)-(n-2003 s) \\
=2003 s-2002 r=2003(r+s)-4005 r
\end{array}
$$

is a multiple of 9.
Since 4005 is a multiple of 9, and 2003 is coprime with 9, $r+s$ must be a multiple of 9.

If $r=s=0$, then $n \geqslant 9 \times 2003$ (because in this case $b_{1}, b_{2}, \cdots, b_{2003}$ are all divisible by 9). If $r \neq 0$, then $r+s=9$. Thus, at least one of $r$ and $s$ is no less than 5. In this case, we get $n \geqslant 5 \times 2002$ or $n \geqslant 5 \times 2003$.
Since $10010=5 \times 2002=4 \times 2002+2002 \times 1$, and 4 and 2002 have the same sum of digits, 10010 is the desired number.