For example, the sum of $4 m$ distinct positive even numbers and $n$ distinct positive odd numbers is 1987. For all such $m$ and $n$, what is the maximum value of $3 m + 4 n$? Please prove your conclusion.

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Solution 1: The maximum value of $3 m+4 n$ is 221.
Let $a_{1}+a_{2}+\cdots+a_{m}+b_{1}+b_{2}+\cdots+b_{n}=1987$, where $a_{i}(i=1, \cdots, m)$ are $m$ distinct positive even numbers, and $b_{j}(j=1, \cdots, n)$ are $n$ distinct positive odd numbers, then $n$ must be odd, and we have
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{m} \geqslant 2+4+\cdots+2 m=m(m+1), \\
b_{1}+b_{2}+\cdots+b_{n} \geqslant 1+3+\cdots+(2 n-1)=n^{2} .
\end{array}
$$

Therefore, $m^{2}+m+n^{2} \leqslant 1987$ (where $n$ is odd). This inequality is equivalent to $\left(m+\frac{1}{2}\right)^{2}+n^{2} \leqslant 1987+\frac{1}{4}$.
Using the Cauchy-Schwarz inequality, we have
$$
\left[3\left(m+\frac{1}{2}\right)+4 n\right]^{2} \leqslant\left(3^{2}+4^{2}\right)\left[\left(m+\frac{1}{2}\right)^{2}+n^{2}\right] \leqslant 5^{2}\left(1987+\frac{1}{4}\right) ,
$$

then $3 m+4 n \leqslant\left[\left(5 \sqrt{1987+\frac{1}{4}}-\frac{3}{2}\right)\right]$, where $[x]$ denotes the integer part of $x$, i.e., $3 m+4 n \leqslant 221$.

On the other hand, when $m=27, n=35$, $m^{2}+m+n^{2} \leqslant 1987$, and $3 m+4 n=221$. Therefore, for $m$ and $n$ satisfying the conditions of the problem, the maximum value of $3 m+4 n$ is 221.
Solution 2: Using trigonometric substitution, from Solution 1 we have
$$
m(m+1)+n^{2} \leqslant 1987(*) \text {, i.e., }\left(m+\frac{1}{2}\right)^{2}+n^{2} \leqslant 1987+\frac{1}{4} \text {. }
$$

Let $m=-\frac{1}{2}+k \cos \theta, n=k \sin \theta\left(k^{2} \leqslant 1987+\frac{1}{4}, \theta \in \mathbb{R}\right)$, then
$$
\begin{aligned}
3 m+4 n & =k(3 \cos \theta+4 \sin \theta)-\frac{3}{2}=5 k \sin (\varphi+\theta)-\frac{3}{2} \\
& \leqslant 5 \sqrt{1987+\frac{1}{4}}-\frac{3}{2}<222 .
\end{aligned}
$$

Thus, the maximum value of $3 m+4 n$ is 221. When $3 m+4 n=221$, solving (*) gives $m=27, n=35$. Therefore, we can take 27 even numbers $2,4,6, \cdots, 50,52,(54+6)$, and 35 odd numbers $1,3, \cdots, 67,69$.