Solve: Represent 1976 as the sum of several positive integers:
$1976=x_{1}+\cdots+x_{k}$, where the number of terms $k$ and each term $x_{1}, \cdots, x_{k}$ are positive integers.

Since $1 \leqslant k \leqslant 1976,1 \leqslant x_{j} \leqslant 1976,1 \leqslant j \leqslant k$, there are only a finite number of such representations, and the corresponding products of positive integers $x_{1} \cdots x_{k}$ also take only a finite number of values, which have an upper bound. For example, it is clear that $1 \leqslant x_{1} \cdots x_{k} \leqslant 1976^{1976}$. Therefore, according to the principle of the largest natural number, there must be a maximum value among these products, denoted as $A$, i.e., there exist positive integers $t$ and $a_{1}, \cdots, a_{t}$, such that for any $k$ and $x_{1}, \cdots, x_{k}$ satisfying equation (1), we have
$$1976=a_{1}+\cdots+a_{t}, \quad A=a_{1} \cdots a_{t} \geqslant x_{1} \cdots x_{k} .$$

Next, we will specifically find such positive integers $t$ and $a_{1}, \cdots, a_{t}$.