For example, the sum of $15 m$ 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.
(CMO - 2 problem)
Prove that if $a_{1}+a_{2}+\cdots+a_{m}+b_{1}+b_{2}+\cdots+b_{n}=1987$, where $a_{i}(1 \leqslant i \leqslant m)$ are distinct positive even numbers, and $b_{j}(1 \leqslant j \leqslant n)$ are distinct positive odd numbers, then $n$ must be odd, and
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{m} \geqslant 2+4+\cdots+2 m=m(m+1), \\
b_{1}+b_{2}+\cdots+b_{n} \geqslant 1+3+\cdots+(2 n-1)=n^{2} .
\end{array}
$$

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

which implies $3 m+4 n \leqslant 5 \sqrt{1987+\frac{1}{4}}-\frac{3}{2}$.
Since $3 m+4 n$ is an integer, we have
$3 m+4 n \leqslant\left[5 \sqrt{1987+\frac{1}{4}}-\frac{3}{2}\right]$, i.e., $3 m+4 n \leqslant 221$.
It is easy to verify that the integer solutions of the equation $3 m+4 n=221$ are of the form
$$
\left\{\begin{array}{l}
m=71-4 k, \\
n=2+3 k(k \text { is an integer). }
\end{array}\right.
$$

Since $n$ is odd, $k$ in (1) must be odd. Let $k=2 t+1, t$ is an integer, then
$$
\left\{\begin{array}{l}
m=67-8 t, \\
n=5+6 t(t \text { is an integer }) .
\end{array}\right.
$$

Since $m^{2}, n^{2} \leqslant 1987$, we have $m, n \leqslant[\sqrt{1987}]=44$.
Substituting into (2), we get $3 \leqslant t \leqslant 6$.
Substituting $t=3,4,5,6$ into (2) respectively, we find that $(m, n)$ can only be $(43,23),(35,29),(27,35)$, and $(19,41)$.
It is not difficult to verify that $(43,23),(35,29),(19,41)$ do not satisfy the relation $m(m+1)+n^{2} \leqslant 1987$.
For $(27,35)$, since $27(27+1)+35^{2}=1981<1987$,
by appropriately selecting 27 positive even numbers and 35 positive odd numbers, we can make their sum exactly 1987.
For example, $2+4+\cdots+54+1+3+\cdots+67+69=1981$,
then $2+4+\cdots+54+1+3+\cdots+67+75=1987$.
In summary, the maximum value of $3 m+4 n$ is 221, and it can only be achieved when $m=27, n=35$.