4. (28th Russian Mathematical Olympiad) Find the smallest positive integer that can be represented as the sum of 2002 positive integers with equal sums of digits, and also as the sum of 2003 positive integers with equal sums of digits.
4. Suppose for the positive integer $n$ there is the expression $n=a_{1}+a_{2}+\cdots+a_{2002}=b_{1}+b_{2}+\cdots+b_{2003}$.

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, denoted as $r(0 \leqslant r \leqslant 8)$. 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,
$(n-2002 r)-(n-2003 s)=2003 s-2002 r=2003(r+s)-4005 r$ is a multiple of 9.
Since 4005 is a multiple of 9 and 2003 is coprime with 9, $r+s$ is 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 $r+s=9$, so 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 the sum of the digits of 4 and 2002 is the same, 10010 is the desired number.