13. There are 2006 distinct complex numbers, such that the product of any two of them (including self-multiplication) is one of these 2006 numbers. Find the sum of these 2006 numbers.
Three, 13. Let there be $n$ distinct complex numbers $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$.
If one of them is 0, without loss of generality, let $\alpha_{n}=0$, then the set
$$
\left\{\alpha_{1}^{2}, \alpha_{1} \alpha_{2}, \cdots, \alpha_{1} \alpha_{n-1}\right\}=\left\{\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n-1}\right\} .
$$

Thus, by $\alpha_{1}^{2} \cdot \alpha_{1} \alpha_{2} \cdots \cdot \alpha_{1} \alpha_{n-1}=\alpha_{1} \cdot \alpha_{2} \cdots \cdot \alpha_{n-1}$, we get $\alpha_{1}^{n-1}=1$.
Similarly, $a_{k}^{n-1}=1(k=1,2, \cdots, n-1)$.
Since $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n-1}$ are distinct complex numbers, we have $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n-1}=0$, which means
$$
\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=0 .
$$

If these $n$ complex numbers are all non-zero, similarly, we get $\alpha_{k}^{n}=1$. Since $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{n}$ are distinct complex numbers, we have $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}=0$