Example 8 Suppose $N$ consecutive positive integers satisfy the following conditions: the sum of the digits of the 1st number is divisible by 1, the sum of the digits of the 2nd number is divisible by 2, $\cdots$. The sum of the digits of the $N$th number is divisible by $N$. Find the maximum possible value of $N$.


Let the $N$ numbers be $a_{1}, a_{2}, \cdots, a_{N}$.
If $N \geqslant 22$, then among $a_{2}, a_{3}, \cdots, a_{21}$, these 20 numbers, at least two numbers have a units digit of 9, and among these two numbers, at least one has a tens digit that is not 9, let this number be $a_{i}$.
Thus, $i \leqslant 21$.
If $i$ is even, then $i \leqslant 20$.
Since $i \mid S\left(a_{i}\right)$, $S\left(a_{i}\right)$ is also even.
Since $S\left(a_{i+2}\right)=S\left(a_{i+1}\right)+1=S\left(a_{i}\right)-7$ is odd, we get $(i+2) \times S\left(a_{i+2}\right)$.
Therefore, $N \leqslant i+1 \leqslant 21$.
If $i$ is odd, then from $(i-1) \mid S\left(a_{i-1}\right)$, we get $S\left(a_{i-1}\right)$ is even, hence $S\left(a_{i+1}\right)$ is odd.
So, $(i+1) \times S\left(a_{i+1}\right)$.
Thus, $N \leqslant i \leqslant 21$.
In summary, $N \leqslant 21$.
Below is an example for 21.
$$
\begin{array}{l}
\text { By }[2,3, \cdots, 11]=9 \times 3080 \text {, we know that we can let } \\
a_{1}=38 \underset{3078 \uparrow}{99 \ldots 989} \text {, } \\
a_{i}=a_{1}+(i-1)(i=2,3, \cdots, 21) . \\
\end{array}
$$

Thus $2 \mid S\left(a_{2}\right)=9 \times 3080+2$,
$$
k \mid S\left(a_{k}\right)=9 \times 3081+k(k=3,4, \cdots, 11) \text {. }
$$

For $k=12,13, \cdots, 21$, $S\left(a_{k}\right)=k$.
Thus $k \mid S\left(a_{k}\right)$, but $22 \times S\left(a_{22}\right)=13$.