Example 2 Given that $x_{1}, x_{2}, \cdots, x_{67}$ are positive integers, and their sum equals 110. Find the maximum value of $x_{1}^{2}+x_{2}^{2}+\cdots+x_{67}^{2}$.
Explanation: Start with any set of 67 positive integers $x_{1}, x_{2}, \cdots, x_{67}$ whose sum is 110. Without loss of generality, assume
$$
x_{1} \leqslant x_{2} \leqslant \cdots \leqslant x_{66} \leqslant x_{67}.
$$

First, freeze $x_{2}, x_{3}, \cdots, x_{66}$ and only study $x_{1}$ and $x_{67}$. Since
$$
\begin{array}{l}
\left(x_{1}-1\right)^{2}+\left(x_{67}+1\right)^{2} \\
=x_{1}^{2}+x_{67}^{2}+2+2\left(x_{67}-x_{1}\right)>x_{1}^{2}+x_{67}^{2},
\end{array}
$$

this indicates that if the smallest number $x_{1}$ is decreased by 1 and the largest number $x_{67}$ is increased by 1 (the sum of the 67 positive integers remains unchanged), their sum of squares increases.
For this reason, we perform the following adjustment:
Each time, 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, unfreeze $x_{2}$ and adjust $x_{2}$, still by decreasing $x_{2}$ by 1 and adding 1 to $x_{67}$ each time, until $x_{2}=1$, thus completing the adjustment for $x_{2}$.

Continue this process, adjusting $x_{3}, x_{4}, \cdots, x_{66}$ step by step, until the set $\left(x_{1}, x_{2}, \cdots, x_{66}, x_{67}\right)$ is adjusted to $(1,1, \cdots, 1,44)$.
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
$$
\underbrace{1^{2}+\cdots+1^{2}}_{66 \text{ times}}+44^{2}=2002.
$$