For example, the sum of $10 m$ 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.
Solution: The maximum value of $3 m+4 n$ is 221.
Since the sum of $m$ distinct positive even numbers $\geqslant 2+4+6+\cdots+2 m=m^{2}+m$, and the sum of $n$ distinct positive odd numbers $\geqslant 1+3+5+\cdots+2 n-1=n^{2}$, thus $m^{2}+m+n^{2} \leqslant 1987$, which means $\left(m+\frac{1}{2}\right)^{2}+n^{2} \leqslant 1987+\frac{1}{4}$.

Also, since $m, n \in N$, the point $(m, n)$ is a lattice point within or on the boundary of the circle $\left(x+\frac{1}{2}\right)^{2}+y^{2}=1987+\frac{1}{4}$ in the first quadrant. Therefore, we can let $m+\frac{1}{2}=\operatorname{tr} \cos \theta, n=t r \sin \theta$, where $0<t \leqslant 1, \theta \in\left(0, \frac{\pi}{2}\right), r=\sqrt{\left(1987+\frac{1}{4}\right.}$.

Thus, $3 m+4 n=3 t r \cos \theta+4 t r \sin \theta-\frac{3}{2} \leqslant(3 \cos \theta+4 \sin \theta) r-\frac{3}{2}$ $\leqslant 5 r-\frac{3}{2}<5 \times 44.6-\frac{3}{2}=221.5$.

On the other hand, we can take the sum of 27 consecutive positive even numbers $2,4,6$, $\cdots 54$ which is 756; and the sum of 35 odd numbers, which includes 34 consecutive positive odd numbers $1,3,5, \cdots$ 67 and the odd number 75, which is 1231. Then the sum of all these numbers is $756+1231=1987$, and $3 \times 27+$ $4 \times 35=221$.
Therefore, the maximum value of $3 m+4 n$ is 221.