16. (12 points) A clothing workshop has 4 teams, A, B, C, and D, with their daily production capacities for shirts and pants as shown in Table 1. Now, it is required to produce matching sets of shirts and pants (one shirt and one pair of pants make a set). How many sets can these 4 teams produce in 7 days?
16. The ratio of shirts to skirts produced daily by groups $A$, $B$, $C$, and $D$ are $\frac{8}{10}$, $\frac{9}{12}$, $\frac{7}{11}$, and $\frac{6}{7}$, respectively, and
$$
\frac{6}{7}>\frac{8}{10}>\frac{9}{12}>\frac{7}{11} \text {. }
$$

Only by having the group with the highest efficiency in producing shirts make shirts, and the group with the highest efficiency in producing skirts make skirts, can the number of sets produced be maximized.

From (1), we know that group $D$ has the highest efficiency in producing shirts, and group $C$ has the highest efficiency in producing skirts. Therefore, let group $A$ work on shirts for $x$ days and on skirts for $(7-x)$ days; group $B$ work on shirts for $y$ days and on skirts for $(7-y)$ days; group $D$ work on shirts for 7 days, and group $C$ work on skirts for 7 days. The total number of shirts produced by the four groups in 7 days is
$$
(6 \times 7 + 8x + 9y) \text{ pieces; }
$$

The total number of skirts produced is
$$
[11 \times 7 + 10(7-x) + 12(7-y)] \text{ pieces. }
$$

According to the problem, we have
$$
42 + 8x + 9y = 77 + 10(7-x) + 12(7-y),
$$

which simplifies to $y = 9 - \frac{6x}{7}$.
Let $\mu = 42 + 8x + 9y = 123 + \frac{2}{7}x$.
Since $0 \leqslant x \leqslant 7$, when $x = 7$, $y = 3$, and $\mu$ reaches its maximum value, i.e., $\mu_{\text{max}} = 125$.