II. (50 points)
Let $a_{i} \equiv 1(\bmod 3), 1 \leqslant i \leqslant m, b_{i} \equiv 2(\bmod 3), 1 \leqslant i \leqslant n, c_{i} \equiv 0(\bmod 3), 1 \leqslant i \leqslant$ $k$, and $a_{i}, b_{i}, c_{i}$ these $m+n+k$ numbers are all distinct. If $\sum_{i=1}^{m} a_{i}+\sum_{i=1}^{n} b_{i}+\sum_{i=1}^{k} c_{i}=2007$, find the maximum value of $4 m+3 n+5 k$.
2. Since $\sum_{i=1}^{m} a_{i} \geqslant 1+4+\cdots+(3 m-2)=\frac{1}{2}(3 m-1) \cdot m$,
$$
\begin{array}{l}
\sum_{i=1}^{n} b_{2} \geqslant 2+5+\cdots+(3 n-1)=\frac{1}{2}(3 n+1) \cdot n, \\
\sum_{i=1}^{k} c_{i} \geqslant 3+6+\cdots+3 k=\frac{1}{2}(3 k+3) \cdot k,
\end{array}
$$

Therefore, $2008 \geqslant \frac{3}{2} \cdot\left(m^{2}-\frac{1}{3} m\right)+\frac{3}{2}\left(n^{2}+\frac{1}{3} n\right)+\frac{3}{2}\left(k^{2}+k\right)$
$$
\begin{array}{c}
\left(m^{2}-\frac{1}{3} m\right)+\left(n^{2}+\frac{1}{3} n\right)+\left(k^{2}+k\right) \leqslant \frac{4014}{3} \\
\left(m-\frac{1}{6}\right)^{2}+\left(n+\frac{1}{6}\right)+\left(k+\frac{1}{2}\right)^{2} \leqslant \frac{48179}{36} \\
\Rightarrow(16+9+25) \cdot \frac{48179}{36} \geqslant\left[\left(4 m-\frac{2}{3}\right)+\left(3 n+\frac{1}{2}\right)+\left(5 k+\frac{5}{2}\right)\right]^{2} \\
=\left(4 m+3 n+5 k+\frac{7}{3}\right)^{2} .
\end{array}
$$

Therefore, $4 m+3 n+5 k \leqslant 256$.
Taking $m=20, n=17, k=25$,
$$
\begin{aligned}
\text { Let } a_{1}, a_{2}, \cdots, a_{20} & =1,4,7 \cdots 58, \\
b_{1}, b_{2}, \cdots, b_{17} & =2,5,8 \cdots 50, \\
c_{1}, c_{2}, \cdots, c_{25} & =3,6,9 \cdots 75,
\end{aligned}
$$

It is verified to be true.