Example 1 (1987 2nd National Mathematical Winter Camp Question) The sum of $m$ distinct positive even numbers and $n$ distinct positive odd numbers is 1987. For all such $m$ and $n$, what is the maximum value of $3m + 4n$? Please prove your conclusion.
Thought analysis: First, find an upper bound for $3 m+4 n$ based on the given conditions, then provide an example to show that this upper bound can be achieved, thus obtaining the maximum value of $3 m+4 n$.

Solution: Let $a_{1}, a_{2}, \cdots, a_{m}$ be distinct positive even numbers, and $b_{1}, b_{2}, \cdots, b_{n}$ be distinct positive odd numbers, such that $a_{1}+a_{2}+\cdots+a_{m}+b_{1}+b_{2}+\cdots+b_{n}=1987$,

At this point, we have: $a_{1}+a_{2}+\cdots+a_{m} \geqslant 2+4+\cdots+2 m=m(m+1)$,
$$
b_{1}+b_{2}+\cdots+b_{n} \geqslant 1+3+\cdots+(2 n-1)=
$$

From (1), (2), and (3), we get $m^{2}+m+n^{2} \leqslant 1987$, thus $\left(m+\frac{1}{2}\right)^{2}+n^{2} \leqslant 1987+\frac{1}{4}$,

From (4) and the Cauchy-Schwarz inequality, we get $3\left(m+\frac{1}{2}\right)+4 n \leqslant \sqrt{3^{2}+4^{2}} \cdot \sqrt{\left(m+\frac{1}{2}\right)^{2}+n^{2}} \leqslant 5 \sqrt{1987+\frac{1}{4}}$,
Since $3 m+4 n$ is an integer, we have $3 m+4 n \leqslant 221$, (5)
On the other hand, when $m=27, n=35$, we have $m^{2}+m+n^{2}=1981<1987$, and $3 m+4 n=221$.
Therefore, the maximum value of $3 m+4 n$ is 221.