(Given: $x_{1}+x_{2}+\cdots+x_{n}=1976, u=x_{1} x_{2} \cdots$ $x_{1976}$. If $u$ takes the maximum value, then $x_{i} \leqslant 4(i=1,2, \cdots, n)$. This is because if some $x_{j}>4$, then replacing $x_{j}$ with 2 and $x_{j}-2$ will increase $u$; similarly, it can be shown that all $x_{i} \geqslant 2$. Since a $x_{i}=4$ can be replaced by two 2s without changing the value of $u$, therefore, $x_{i}=2$ or $3(i=1,2, \cdots, n)$. Furthermore, from $3^{2}>2^{3}$ and $1976=658 \times 3+2$, we get $u_{\max }=2 \times 3^{658}$.)