13. Given $n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying
$$
x_{1}+x_{2}+\cdots+x_{n}=2008 .
$$

Find the maximum value of the product $x_{1} x_{2} \cdots x_{n}$ of these $n$ positive integers.
13. Let the maximum value of $x_{1} x_{2} \cdots x_{n}$ be $M$.

Since $x_{1}+x_{2}+\cdots+x_{n}=2008$, it is clear that each $x_{i} (i=1,2, \cdots, n)$ in $M$ is greater than 1.

If there is an $x_{i} \geqslant 4$, we can split $x_{i}$ into $x_{i}-2$ and 2, and examine their product:
$$
\begin{array}{l}
\left(x_{i}-2\right) \times 2=2 x_{i}-4 \\
=x_{i}+\left(x_{i}-4\right) \geqslant x_{i},
\end{array}
$$

Thus, by splitting any positive integer $x_{i}$ that is greater than or equal to 4 into $x_{i}-2$ and 2, the sum remains unchanged, but the product increases.

Therefore, in the maximum value $M=x_{1} x_{2} \cdots x_{n}$, there cannot be any positive integers greater than or equal to 4.

Hence, $x_{i}=2$ or $x_{i}=3$, meaning $M$ can be written in the form $2^{p} \cdot 3^{q}$.

Furthermore, since $2+2+2=3+3$, but $2^{3}<3^{2}$, replacing 3 twos with 2 threes in the product increases the product, so $p \leqslant 2$.
$$
\text { Also, } 2008=3 \times 668+2+2 \text {, hence } M=2^{2} \times 3^{668} \text {. }
$$