9. Given $n(n>1)$ integers (which can be the same) $a_{1}$, $a_{2}, \cdots, a_{n}$ satisfy
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=2007 .
$$

Then the minimum value of $n$ is $\qquad$
9. 5 .

Given $a_{1} a_{2} \cdots a_{n}=2007$, we know that $a_{1}, a_{2}, \cdots, a_{n}$ are all odd numbers. Also, $a_{1}+a_{2}+\cdots+a_{n}=2007$ is an odd number, so $n$ is odd.

If $n=3$, i.e., $a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3}=2007$, without loss of generality, assume $a_{1} \geqslant a_{2} \geqslant a_{3}$, then
$$
a_{1} \geqslant \frac{a_{1}+a_{2}+a_{3}}{3}=669, a_{2} a_{3} \leqslant \frac{2007}{a_{1}} \leqslant 3 \text {. }
$$

If $a_{1}=669$, then $a_{2} a_{3}=3$. Thus,
$a_{2}+a_{3} \leqslant 4, a_{1}+a_{2}+a_{3} \leqslant 673669$, only $a_{1}=2007, a_{2} a_{3}=1$ and $a_{2}+a_{3}$ $=0$, which is impossible.
Therefore, $n \geqslant 5$.
$$
\begin{array}{l}
\text { Also, } 2007+1+1+(-1)+(-1) \\
=2007 \times 1 \times 1 \times(-1) \times(-1)=2007,
\end{array}
$$

Thus, the minimum value of $n$ is 5.