(6) 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.
To make $s=3m+4n$ as large as possible, we need to make the $m$ even numbers $a_1, \cdots, a_m$ and the $n$ odd numbers $b_1, \cdots, b_n$ as small as possible. Therefore, we can consider the inequality:
$$
(2+4+\cdots+2m) + (1+3+\cdots+(2n-1)) \leq 1987
$$

Equation (1) is:
$$
m(m+1) + n^2 \leq 1987
$$

Equation (2) represents a disk in the $mn$ plane:
$$
\left(m+\frac{1}{2}\right)^2 + n^2 \leq \frac{7949}{4}
$$

All pairs $(m, n)$ that satisfy the conditions, with these coordinates, fall within the intersection of the disk (3) and the first quadrant (including the coordinate axes), as shown in Figure 1.

Consider a set of parallel lines: $3m + 4n = s$ (where $s$ is a positive integer). The maximum value of $3m + 4n$ should be the maximum value of $s$ for which the line $3m + 4n = s$ intersects the disk (3) at integer points.

If $s$ is not restricted to positive integers, the maximum value of $s$ for which the set of parallel lines $3m + 4n = s$ intersects the disk (3) occurs when $3m + 4n = s$ is tangent to the disk (3). At this point, the tangent point $P(m, n)$ satisfies:
$$
-\frac{3}{4} = -\frac{m+\frac{1}{2}}{n} \Rightarrow 3n = 4m + 2
$$

Substituting into:
$$
m(m+1) + n^2 = 1987
$$

We get:
$$
\begin{aligned}
m & = \frac{3}{10} \sqrt{7949} - \frac{1}{2}, \quad n = \frac{2}{5} \sqrt{7949} \\
s & = 3m + 4n = \frac{5}{2} \sqrt{7949} - \frac{3}{2}
\end{aligned}
$$

Since:
$$
89^2 < 7949 < 89.16^2
$$

We have:
$$
221 < s < 221.4
$$

Considering $s$ is a positive integer, the maximum value of $s$ is less than or equal to 221. Moving the line $l$ parallel to the left until it hits the first lattice point $Q(m, n)$ (within the disk (3)), then $s = 3m + 4n$ is the maximum value. Since $n$ is clearly odd, we should rephrase this as "hitting the first lattice point $Q(m, n)$ where $n$ is odd, and this point's $y$-coordinate $n$ should be as close as possible to $\frac{2}{5} \sqrt{7949} = 35.6 \cdots$, so we consider $n = 35$ or $n = 37$. The former makes $m(m+1) \leq 1987 - n^2 = 762$, with the maximum value of $m$ being 27, and the maximum value of $s$ being 221. The latter can be similarly calculated to give a maximum value of $s$ as 220, so we should take the former. An actual construction can be (not unique):
$$
\begin{array}{c}
a_1 = 2, a_2 = 4, \cdots, a_{26} = 52, a_{27} = 60 \\
m = 27, b_1 = 1, b_2 = 3, \cdots, b_{35} = 69 \\
n = 35
\end{array}
$$

Then we get $a_1 + a_2 + \cdots + a_{27} + b_1 + \cdots + b_{35} = 1987$
Similarly, $s = 3m + 4n = 221$ is the maximum value.