$7 \cdot 56$ If there are $n$ integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying the following equation
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} a_{2} \cdots a_{n}=1990 \text {. }
$$

Find the smallest natural number $n(n \geqslant 2)$.
(16th All-Russian Mathematical Olympiad, 1990)
［Solution］Since $a_{1} a_{2} \cdots a_{n}=1990$ is divisible by 2 but not by 4, there is only one even number among $a_{1}, a_{2}, \cdots, a_{n}$.
Also, from $a_{1}+a_{2}+\cdots+a_{n}=1990$ and the fact that there is only one even number in $a_{i}$, it follows that $n$ is

an odd number.
Next, we prove that $n \neq 3$.
Without loss of generality, assume $a_{1} \geqslant a_{2} \geqslant a_{3}$.
From $a_{1} \geqslant \frac{1990}{3}$, we know that $a_{1}$ is either 1990 or 995.
If $a_{1}=1990$, then $\left|a_{2}\right|=\left|a_{3}\right|=1$, and $a_{2}$ and $a_{3}$ have the same sign, but in this case
$$
a_{1}+a_{2}+a_{3} \neq 1990 \text {. }
$$

If $a_{1}=995$, then $\left|a_{2}\right| \leqslant 2,\left|a_{3}\right| \leqslant 2$, thus
$$
a_{1}+a_{2}+a_{3}<1990 \text {. }
$$

Therefore, $n \neq 3$.
If $n=5$, take $a_{1}=1990, a_{2}=a_{3}=1, a_{4}=a_{5}=-1$, then we have
$$
\begin{array}{c}
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=1990, \\
a_{1} a_{2} a_{3} a_{4} a_{5}=1990 .
\end{array}
$$

This satisfies the conditions.
Thus, the smallest natural number $n=5$.