Tokarev S.i.

Find the maximum number $N$ for which there exist $N$ consecutive natural numbers such that the sum of the digits of the first number is divisible by 1, the sum of the digits of the second number is divisible by 2, the sum of the digits of the third number is divisible by 3, ..., and the sum of the digits of the $N$-th number is divisible by $N$.
Evaluation. Let's denote the sum of the digits of a number $a$ by $S(a)$. Suppose there exist 22 consecutive numbers $a_{1}, \ldots, a_{22}$, satisfying the conditions of the problem. Then the numbers $S\left(a_{2}\right), S\left(a_{12}\right)$, and $S\left(a_{22}\right)$ are even. On the other hand, if the second-to-last digit of the number $a_{2}$ is not a nine, then $S\left(a_{12}\right)=S\left(a_{2}+10\right)=S\left(a_{2}\right)+1$. Contradiction. If, however, the second-to-last digit of the number $a_{2}$ is a nine, then the second-to-last digit of the number $a_{12}$ is 0 and $S\left(a_{22}\right)=S\left(a_{12}+10\right)$ $=S\left(a_{12}\right)+1$. Again, a contradiction.

Example with 21 numbers. Let $m=5 \cdot 7 \cdot 8 \cdot 11$. Consider the consecutive numbers:

3
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$38 \underbrace{0 . .09}_{m-1}$. The sums of the digits of these numbers are $9 m+10, 9 m+2, 9 m+3, \ldots, 9 m+11, 12, \ldots, 21$.