Tokarev S.I.

Find the smallest natural number that can be represented as the sum of 2002 natural addends with the same sum of digits and as the sum of 2003 natural addends with the same sum of digits.

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Let for a natural number $n$ the following representations hold: $n=a_{1}+\ldots+a_{2002}=b_{1}+\ldots+b_{2003}$.

We will use the fact that each of the numbers $a_{1}, a_{2002}$ gives the same remainder when divided by 9 as the sum of its digits; let this remainder be $r(0 \leq r \leq 8)$, and the corresponding remainder for the numbers $b_{1}, \ldots, b_{2003}$ be $s$ ($0 \leq s \leq 8$). Then the numbers $n-2002 r$ and $n-2003 s$ are multiples of 9, and therefore, the number $(n-2002 r)-(n-2003 s)=2003 s-2002 r=2003(r+s)-4005 r$ is also a multiple of 9. The number $4005 r$ is also a multiple of 9, and the number 2003 is coprime with 9; hence, the number $r+s$ is a multiple of 9. If in this case $r=s=0$, then $n \geq 9 \cdot 2003$ (since $b_{1}, \ldots, b_{2003}$ are divisible by 9). If $r \neq 0$, then $r+s=9$, and therefore $r \geq 5$ or $s \geq 5$ for the number $n$, the inequalities $n \geq 5 \cdot 2002$ or $n \geq 5 \cdot 2003$ are obtained, respectively. Since $10010=5 \cdot 2002=4 \cdot 2002+$

$2002 \cdot 1$, and the numbers 4 and 2002 have the same sum of digits, the number 10010 is the desired one.

## Answer

10010.