14. A. Let $a_{1}, a_{2}, \cdots, a_{n}$ be pairwise distinct positive integers, and satisfy $a_{1}+a_{2}+\cdots+a_{n}=2014$, where $n$ is some integer greater than 1. Let the greatest common divisor of $a_{1}, a_{2}, \cdots, a_{n}$ be $d$. For all $n$ and $a_{1}, a_{2}, \cdots, a_{n}$ that satisfy the above conditions, find the maximum value of $n d$.
14. A. Let $a_{i}=b_{i} d(i=1,2, \cdots, n)$, where $b_{1}, b_{2}, \cdots, b_{n}(n \geqslant 2)$ are distinct positive integers, and without loss of generality, assume $1 \leqslant b_{1}<b_{2}<\cdots<b_{n}$.
Thus, $b_{i} \geqslant i$.
Therefore, $2014=a_{1}+a_{2}+\cdots+a_{n}$
$$
\begin{array}{l}
=\left(b_{1}+b_{2}+\cdots+b_{n}\right) d \\
\geqslant(1+2+\cdots+n) d=\frac{n(n+1)}{2} d .
\end{array}
$$

Since $n \geqslant 2$, from equation (1) we get
$$
d \leqslant \frac{2014}{3} \text {. }
$$

Also, since $d \mid 2014$, and $2014=2 \times 19 \times 53$, the possible values of $d$ are $1, 2, 19, 38, 53, 106$.
If $d=106$, then from equation (1) we know
$$
n(n+1) \leqslant \frac{2 \times 2014}{d}=38 \text {. }
$$

Since $n$ is a positive integer, we have $n \leqslant 5$, and in this case,
$$
n d \leqslant 5 \times 106=530 \text {. }
$$

If $d<106$, then $d \leqslant 53$, and from equation (1) we get
$$
\begin{array}{l}
n d=\sqrt{n^{2} d^{2}} \leqslant \sqrt{\frac{n(n+1)}{2} d \cdot 2 d} \\
\leqslant \sqrt{2014 \times 2 \times 53}=106 \sqrt{19}<530 .
\end{array}
$$

Therefore, we always have $n d \leqslant 530$.
When $n=5, a_{k}=106 k(k=1,2, \cdots, 5)$, $n d$ reaches its maximum value of 530.