11. (20 points) Find the smallest integer \( n (n > 1) \), such that there exist \( n \) integers \( a_{1}, a_{2}, \cdots, a_{n} \) (allowing repetition) satisfying
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2013 .
$$
12. (20 points) Let positive integers \( a, b, c, d \) satisfy
$$
a^{2}=c(d+13), b^{2}=c(d-13) \text {. }
$$

Find all possible values of \( d \).
11. Since $a_{1} a_{2} \cdots a_{n}=2013$, it follows that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. From $a_{1}+a_{2}+\cdots+a_{n}=2013$ being odd, we know that $n$ is odd (otherwise, the sum of an even number of odd numbers should be even, which is a contradiction).
If $n=3$, then $a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3}=2013$.
Assume without loss of generality that $a_{1} \geqslant a_{2} \geqslant a_{3}$. Then
$a_{1} \geqslant \frac{a_{1}+a_{2}+a_{3}}{3}=\frac{2013}{3}=671$
$\Rightarrow a_{2} a_{3}=\frac{2013}{a_{1}} \leqslant \frac{2013}{671}=3$.
If $a_{1}=671$, then $a_{2} a_{3}=3$. Thus, $a_{2}+a_{3} \leqslant 4$.
Hence $a_{1}+a_{2}+a_{3} \leqslant 671+4=675671$, at this point, $a_{2} a_{3}<3$.
Therefore, $a_{2} a_{3}=1, a_{1}=2013$, which contradicts $a_{1}+a_{2}+a_{3}=2013$.
When $n=5$, there exist $-1, -1, 1, 1, 2013$ satisfying
$(-1)+(-1)+1+1+2013$
$=(-1) \times(-1) \times 1 \times 1 \times 2013=2013$.
Thus, the minimum value of $n$ is 5.
12. From the given, we have $\frac{a^{2}}{b^{2}}=\frac{c(d+13)}{c(d-13)}=\frac{d+13}{d-13}$.
Let $(a, b)=t, a_{1}=\frac{a}{t}, b_{1}=\frac{b}{t}$. Then
$\frac{a_{1}^{2}}{b_{1}^{2}}=\frac{a^{2}}{b^{2}}=\frac{d+13}{d-13}$.
Since $\left(a_{1}, b_{1}\right)=1$, we know $\left(a_{1}^{2}, b_{1}^{2}\right)=1$, meaning the left side of equation (1) is a reduced fraction. Assume without loss of generality that
$d+13=k a_{1}^{2}, d-13=k b_{1}^{2}\left(k \in \mathbf{Z}_{+}\right)$.
Thus, $k\left(a_{1}^{2}-b_{1}^{2}\right)=26$, i.e., $a_{1}^{2}-b_{1}^{2}=\frac{26}{k}$ is a positive divisor of 26 and can be decomposed into the product of two unequal positive integers $a_{1}+b_{1}$ and $a_{1}-b_{1}$ with the same parity, hence $\frac{26}{k}=13$.
Then $a_{1}+b_{1}=13, a_{1}-b_{1}=1, k=2$. Thus, $a_{1}=7$.
Furthermore, $d=k a_{1}^{2}-13=2 \times 7^{2}-13=85$.
Upon verification, when $(a, b, c, d)=(14,12,2,85)$, it satisfies the conditions. Therefore, the value of $d$ is 85.