$\therefore 、 m$ 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.

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 $3 m+4 n$? Please prove your conclusion.
The maximum value of $3 m+4 n$ is 221. The proof is as follows: Let $a_{1}+\cdots+a_{n}+b_{1}+\cdots+b_{n}=1987$. 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, $u$ is always an odd number, and
$$
\begin{array}{l}
a_{1}+\cdots+a_{\mathrm{m}} \geqslant 2+4+\cdots+2 m=m(m+1), \\
b_{1}+\cdots+b_{n} \geqslant 1+3+\cdots+(2 n-1)=n^{2} . \\
\therefore m^{2}+m+n^{2} \leqslant 1987 \text {. (where } n \text { is odd) }
\end{array}
$$

This inequality is equivalent to $\left(m+\frac{1}{2}\right)^{2}+n^{2} \leqslant 1987+\frac{1}{4}$. By the Cauchy-Schwarz inequality,
$$
\begin{array}{c}
3\left(m+\frac{1}{2}\right)+4 n \leqslant \sqrt{3^{2}+4^{2}} \\
\cdot \sqrt{\left(m+\frac{1}{2}\right)^{2}+n^{2}} \\
\leqslant 5 \sqrt{1987+\frac{1}{4}} \\
\therefore \quad 3 m+4 n \leqslant\left[\left(5 \cdot \sqrt{1987+\frac{1}{4}}-\frac{3}{2}\right)\right] .
\end{array}
$$

where $[x]$ denotes the integer part of $x$.
Thus, $3 m+4 n \leqslant 221$.
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 proposition, the maximum value of $3 m+4 n$ is 221.