2 - 83 The sum of several positive integers is 1976. Find the maximum value of the product of these positive integers.

Several positive integers add up to 1976. Find the maximum value of the product of these positive integers.
[Solution] Let these numbers be $a_{1}, \cdots, a_{n}$, then $\square$
$$a_{1}+\cdots+a_{n}=1976$$

Assume without loss of generality that $a_{i}<4(1 \leqslant i \leqslant n)$, 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 $2 \times 3^{658}$.