14.5.23 ** If for a natural number $n(\geqslant 2)$, there are integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying
$$
a_{1}+a_{2}+\cdots+a_{n}=a_{1} \cdot a_{2} \cdot \cdots \cdot a_{n}=1990 \text {, }
$$

find the minimum value of $n$.
Since $1990=a_{1} \cdot a_{2} \cdots \cdots a_{n}$ is divisible by 2 but not by 4, one of $a_{1}, a_{2}, \cdots, a_{n}$ is even, and $n-1$ are odd. Also, $a_{1}+a_{2}+\cdots+a_{n}=1990$, so $n$ must be odd. If $n-3$, without loss of generality, let $a_{1} \geqslant a_{2} \geqslant a_{3}$, then $a_{1} \geqslant \frac{1990}{3}$. Since $a_{1}, a_{2}$ are integers, and $a_{1} \cdot a_{2} \cdot a_{3}=$ 1990, so $a_{1}=1990$ or 995. When $a_{1}=1990$, $\left|a_{2}\right|=\left|a_{3}\right|=1$, and $a_{2}$ and $a_{3}$ must have the same sign, in this case $a_{1}+a_{2}+a_{3} \neq 1990$. If $a_{1}=995$, then $\left|a_{2}\right| \leqslant 2,\left|a_{3}\right| \leqslant 2, a_{1}+a_{2}+a_{3}<$ 1990. Therefore, $n \neq 3$. Also, $a_{1}=1990, a_{2}=a_{3}=1, a_{4}=a_{5}=-1$ satisfies the conditions, so $n=5$.