N17 (18-4, USA) The sum of several positive integers is 1976. Find the maximum value of the product of these positive integers.

The sum of several positive integers is 1976, find the maximum value of the product of these positive integers.
Solution: Since there are only a finite number of representations of 1976 as a sum of positive integers, there are also only a finite number of corresponding products, and thus there must be a maximum one. Let the positive integers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy the condition:
$$
a_{1}+a_{2}+\cdots+a_{n}=1976,
$$

and the product
$$
p=a_{1} a_{2} \cdots a_{n}
$$

is maximized. At this point, $a_{i}(i=1,2, \cdots, n)$ must have the following properties:
(1) $a_{i} \leqslant 4(i=1,2, \cdots, n)$. Otherwise, if some $a_{i}>4$, decompose $a_{i}$ into $a_{i}-2$ and 2, keeping the rest unchanged. Since
$$
2\left(a_{i}-2\right)=a_{i}+\left(a_{i}-4\right)>a_{i},
$$

this contradicts the maximality of $p$.
(2) $a_{i} \neq 1(i=1,2, \cdots, n)$. Otherwise, if $a_{1}=1$, then
$$
\left(1+a_{2}\right)+a_{3}+\cdots+a_{n}=1976,
$$

and
$$
\left(1+a_{2}\right) a_{3} \cdots a_{n}>a_{1} a_{2} \cdots a_{n}=p,
$$

which also contradicts the maximality of $p$.
(3) If $a_{i}(i=1,2, \cdots, n)$ contains 4, it can be replaced by $2+2$ without changing the value of the product $p$.
From the above (1) to (3), we know that the maximum product $p$ should have the form
$$
p=2^{r} \cdot 3^{s}
$$

and $r<3$, because if $r \geqslant 3$, then by $2+2+2=3+3$, and $2^{3}<3^{2}$, i.e., replacing 3 twos with 2 threes increases the product. Hence, $r<3$.
By $1976=3 \times 658+2$

we know that the maximum product is $2 \cdot 3^{658}$.