$2 \cdot 83$ The sum of several positive integers is 1976, find the maximum value of the product of these positive integers.

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[Solution] Let these numbers be $a_{1}, \cdots, a_{n}$, then
$$
a_{1}+\cdots+a_{n}=1976 \text {. }
$$

Without loss of generality, assume $a_{i}<4(1 \leqslant i \leqslant n)$, because when $a_{i} \geqslant 4$, $a_{i} \leqslant 2\left(a_{i}-2\right)$, so replacing $a_{i}$ with the two numbers 2 and $a_{i}-2$ will not decrease the product.

Also note that $2 \times 2 \times 2 < 3 \times 3$, so we only need to consider $2^{a} \cdot 3^{b}$, where $a=0,1,2$, and $2a + 3b = 1976$.
From this, we get $a=1, b=658$,
thus the maximum value sought is $2 \times 3^{658}$.