$2 \cdot 93 m$ distinct positive even numbers and $n$ distinct positive odd numbers have a total sum of 1987. For all such $m$ and $n$, what is the maximum value of $3 m+4 n$? Please prove your conclusion.
[Proof] Let $a_{1}+a_{2}+\cdots+a_{m}+b_{1}+b_{2}+\cdots b_{n}=1987$. Here, $a_{i}(1 \leqslant i \leqslant m)$ are distinct positive even numbers, and $b_{j}(1 \leqslant j \leqslant n)$ are distinct positive odd numbers.
Clearly, $n$ 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,
$$
m^{2}+m+n^{2} \leqslant 1987 \text {, where } n \text { is odd. }
$$

This inequality is equivalent to
$$
\left(m+\frac{1}{2}\right)^{2}+n^{2} \leqslant 1987+\frac{1}{4} .
$$

By the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
3\left(m+\frac{1}{2}\right)+4 n \leqslant \sqrt{3^{2}+4^{2}} \sqrt{\left(m+\frac{1}{2}\right)^{2}+n^{2}} \\
\leqslant 5 \sqrt{1987+\frac{1}{4}} \\
3 m+4 n \leqslant 5 \sqrt{1987+\frac{1}{4}}-\frac{3}{2} .
\end{array}
$$

Since $3 m+4 n$ is an integer, we have
$$
3 m+4 n \leqslant\left[5 \sqrt{1987+\frac{1}{4}}-\frac{3}{2}\right] \text {, }
$$

which means $3 m+4 n \leqslant 221$.
It is easy to prove that the integer solutions of the equation $3 m+4 n=221$ are of the form
$$
\left\{\begin{array}{l}
m=71-4 k \\
n=2+3 k
\end{array} \quad(k \text { is an integer), }\right.
$$

Since $n$ is odd, $k$ in (1) must be odd. Let $k=2 t+1, t$ is an integer,
then $\left\{\begin{array}{l}m=67-8 t \\ n=5+6 t\end{array} \quad\right.$ ( $t$ is an integer),
Since $m^{2}, n^{2} \leqslant 1987$, we have
$$
m, n \leqslant[\sqrt{1987}]=44 .
$$

Substituting into (2) yields
$$
3 \leqslant t \leqslant 6 .
$$

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 inequality
$$
m(m+1)+n^{2} \leqslant 1987 \text {. }
$$

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, from
$$
2+4+\cdots+54+1+3+\cdots+67+69=1981,
$$

then $2+4+\cdots+54+1+3+\cdots+67+75=1987$.
In summary, the maximum value of $3 m+4 n$ is 221, and this maximum value can only be achieved when $m=27, n=35$.