14. A. $n$ positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the following conditions:
$$
1=a_{1}<a_{2}<\cdots<a_{n}=2009,
$$

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 $n$.
14. A. Let in $a_{1}, a_{2}, \cdots, a_{n}$, after removing $a_{i}(i=1$, $2, \cdots, n)$, the arithmetic mean of the remaining $n-1$ numbers is a positive integer $b_{i}$, i.e.,
$$
b_{i}=\frac{\left(a_{1}+a_{2}+\cdots+a_{n}\right)-a_{i}}{n-1} .
$$

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

Thus, $(n-1) \mid\left(a_{j}-a_{i}\right)$.
Since $b_{1}-b_{n}=\frac{a_{n}-a_{1}}{n-1}=\frac{2008}{n-1}$ is a positive integer, then $(n-1) \mid 2^{3} \times 251$.
Also, $a_{n}-1$
$$
\begin{array}{l}
=\left(a_{n}-a_{n-1}\right)+\left(a_{n-1}-a_{n-2}\right)+\cdots+\left(a_{2}-a_{1}\right) \\
\geqslant(n-1)+(n-1)+\cdots+(n-1) \\
=(n-1)^{2},
\end{array}
$$

then $(n-1)^{2} \leqslant 2008$.
Thus, $n \leqslant 45$.
Combining $(n-1) \mid 2^{3} \times 251$, we know $n \leqslant 9$.
On the other hand, let
$$
\begin{array}{l}
a_{1}=8 \times 0+1, a_{2}=8 \times 1+1, \\
a_{3}=8 \times 2+1, \cdots \cdots \\
a_{8}=8 \times 7+1, a_{9}=8 \times 251+1 .
\end{array}
$$

Then these 9 numbers satisfy the problem's requirements.
In summary, the maximum value of $n$ is 9.