$3 \cdot 27$ When the natural number $n \geqslant 2$ is the smallest, find integers $a_{1}, a_{2}, \cdots, a_{n}$, such that the following equation $a_{1}+a_{2}+\cdots+a_{n}=a_{1} \cdot a_{2} \cdot \cdots a_{n}=1990$ holds.
[Solution] First, we prove that the minimum value of $n$ is 5. The number 1990 can only be divided by 2 but not by 4, so there is only one even number in $a_{1} \cdot a_{2} \cdots a_{n}$. Therefore, in the sum $a_{1}+a_{2}+\cdots+a_{n} = 1990$, there must be an even number of odd addends, which means $n$ is odd.

If $a_{1}+a_{2}+a_{3}=a_{1} a_{2} a_{3}=1990$, without loss of generality, assume $a_{1} \geqslant a_{2} \geqslant a_{3}$, then $a_{1} \geqslant \frac{1990}{3}$, so $a_{1}$ is 1990 or 995.

If $a_{1}=1990$, then $\left|a_{2}\right|=\left|a_{3}\right|=1$ and $a_{2}$ and $a_{3}$ should have the same sign, clearly $a_{1}+a_{2}+a_{3} \neq 1990$.

If $a_{1}=995$, then $\left|a_{2}\right| \leqslant 2,\left|a_{3}\right| \leqslant 2$, so $a_{1}+a_{2}+a_{3}<1990$, which shows that $n$ cannot be 3.
$$\begin{aligned}
\text{Let } a_{1}=1990, a_{2}=a_{3}=1, a_{4}=a_{5}=-1, \text{ then we have} \\
\\
a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=a_{1} a_{2} a_{3} a_{4} a_{5}=1990 .
\end{aligned}$$

Therefore, the minimum value of $n$ is 5. The required integers are $\{1990,1,1,-1,-1\}$.