For example, $8 n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$

and the arithmetic mean of any $n-1$ different numbers among $a_{1}, a_{2}, \cdots, a_{n}$ is a positive integer. Find the maximum value of $n$. [4]
(2009, "Mathematics Weekly Cup" National Junior High School Mathematics Competition
Let $a_{1}, a_{2}, \cdots, a_{n}$ be such that removing $a_{i} (i=1, 2, \cdots, n)$ leaves the arithmetic mean of the remaining $n-1$ numbers as a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$

Thus, for any $1 \leqslant i<j \leqslant n$, we have
$$
b_{i}-b_{j}=\frac{a_{j}-a_{i}}{n-1} \text {. }
$$

Therefore, $(n-1) \mid\left(a_{j}-a_{i}\right)$.
Since $b_{1}-b_{n}=\frac{a_{n}-a_{1}}{n-1}=\frac{2008}{n-1}$ is a positive integer, we have
$$
(n-1) \mid\left(2^{3} \times 251\right) \text {. }
$$

Also, $a_{n}-1=\sum_{i=1}^{n-1}\left(a_{i+1}-a_{i}\right)$
$$
\geqslant \sum_{i=1}^{n-1}(n-1)=(n-1)^{2} \text {, }
$$

so $(n-1)^{2} \leqslant 2008$, which implies $n \leqslant 45$.
Combining with equation (1), we get $n \leqslant 9$.
On the other hand, let
$$
a_{i+1}=8 i+1(i=0,1, \cdots, 7), a_{9}=8 \times 251+1 \text {. }
$$

Then these nine numbers satisfy the given conditions.
Hence, the maximum value of $n$ is 9.