Example 7 Given $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying: $1<a_{1}<a_{2} \cdots<a_{n}<2009$, and the arithmetic mean of any $n-1$ of these numbers is a positive integer. Find the maximum value of $n$.

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Let $a_{1}, a_{2}, \cdots, a_{n}$ be such that the arithmetic mean of the $n-1$ numbers after removing $a_{i}$ is $b_{i}(i=1,2, \cdots, n)$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$

Thus, for $1 \leqslant i<j \leqslant n$, 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 know $(n-1) \mid 2^{3} \times 251$.
By $a_{n}-a_{1}$
$$
\begin{array}{l}
=\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{2}-a_{1}\right) \\
\geqslant(n-1)+(n-1)+\cdots+(n-1) \\
=(n-1)^{2},
\end{array}
$$

we get $(n-1)^{2} \leqslant 2008$.
Thus, $n \leqslant 45$.
Since $(n-1) \mid 2^{3} \times 251$, we have $n \leqslant 9$.
Below, we construct an example to show that $n=9$ can be achieved.
Let $a_{1}=8 \times 0+1, a_{2}=8 \times 1+1$, ......
$$
a_{8}=8 \times 7+1, a_{9}=8 \times 251+1,
$$

These 9 numbers satisfy the given conditions.
In summary, the maximum value of $n$ is 9.