[Example 3.3.3] Given that $x_{1}, x_{2}, \cdots, x_{67}$ are positive integers, and
$$
x_{1}+x_{2}+\cdots+x_{67}=110 \text {. }
$$

Find the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{67}^{2}$.
We start with any set of 67 positive integers $x_{1}, x_{2}, \cdots, x_{67}$ whose sum is 110, and assume
$$
x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{66} \leqslant x_{67} .
$$

First, we freeze $x_{2}, x_{3}, \cdots, x_{66}$ and only study $x_{1}$ and $x_{67}$. By $x_{1} \leqslant x_{63}$, and noting that
$$
\begin{aligned}
\left(x_{1}-1\right)^{2}+\left(x_{67}+1\right)^{2} & =x_{1}^{2}+x_{67}^{2}+2+2\left(x_{67}-x_{1}\right) \\
& >x_{1}^{2}+x_{67}^{2} .
\end{aligned}
$$

The above inequality shows that if we decrease the smallest number $x_{1}$ by 1 and increase the largest number $x_{67}$ by 1 (keeping the sum of the 67 positive integers unchanged), their sum of squares increases.
For this reason, we perform such an adjustment:
Each time, we decrease $x_{1}$ by 1 and add the 1 to $x_{67}$, until $x_{1}=1$, thus completing the adjustment for $x_{1}$.
The result of this adjustment is that the sum of the 67 positive integers remains 110, while the sum of squares increases after the adjustment compared to before.
Next, we unfreeze $x_{2}$ and adjust $x_{2}$, still decreasing $x_{2}$ by 1 each time and adding 1 to $x_{67}$, until $x_{2}=1$, thus completing the adjustment for $x_{2}$.
We proceed to adjust $x_{3}, x_{4}, \cdots, x_{66}$ step by step in this manner, until we adjust
$$
\begin{array}{c}
\left(x_{1}, x_{2}, \cdots, x_{66}, x_{67}\right) \\
(1,1, \cdots, 1,44) .
\end{array}
$$

At this point, since
$$
1+1+\cdots+1+44=66 \times 1+44=110,
$$

and each adjustment increases the sum of squares, the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{67}^{2}$ is
$$
\underbrace{1^{2}+1^{2}+\cdots+1^{2}}_{66 \text { terms }}+44^{2}=2002 .
$$