2.51 Given that the sum of several positive integers is 1976, find the maximum value of their product.

 untranslated text:
已知若干个正整数之和为 1976 , 求其积的最大值.
[Solution] Since there are only a finite number of different sets of positive integers whose sum is 1976, the maximum value we seek must exist.

Let $x_{1}, x_{2}, \cdots, x_{n}$ be positive integers such that $x_{1}+x_{2}+\cdots+x_{n}=1976$, and let their product $P = x_{1} x_{2} \cdots x_{n}$ be maximized, then we have:
(1) $x_{i} \leqslant 4, i=1,2, \cdots, n$. If there is an $x_{i}>4$, then since $2+(x_{i}-2)=x_{i}$ and $2(x_{i}-2)=2x_{i}-4>x_{i}$, replacing $x_{i}$ with 2 and $x_{i}-2$ will increase the value of the product $P$, leading to a contradiction.
(2) $x_{i} \geqslant 2, i=1,2, \cdots, n$. If there is an $x_{i}=1$, then $x_{i} x_{j}=x_{j}<x_{i}+x_{j}$, so replacing $x_{i}$ and $x_{j}$ with $x_{i}+x_{j}$ will increase the product $P$, leading to a contradiction.
(3) Since $4=2+2=2 \times 2$, if there is an $x_{i}=4$, it can be replaced by two 2s. This way, all $x_{i}$ will be 2 or 3.
(4) From the above arguments, we can assume $P=2^{r} \cdot 3^{s}$, where $r$ and $s$ are non-negative integers. Since $2+2+2=3+3$ and $2^{3}<3^{2}$, it must be that $r<3$. Also, since $1976=658 \times 3+2$, we know that $r=1, s=658$. Therefore, the maximum value we seek is $P=2 \times 3^{658}$.