There are $161 n$ singers participating in the performance of an art festival, and the organizers have arranged $m$ performances for them. It is required that each performance is presented by 4 of the singers, and the number of times any two of the $n$ singers perform together is the same. Please design a scheme so that the total number of performances $m$ is minimized. When
(1) $n=6$,
(2) $n=7$,
(3) $n=9$,

find the minimum value of $m$ for each case.
Solution: Let the number of times each pair of singers performs together be $r$.
(1) $n=6$. At this time, 6 singers can form $\mathrm{C}_{6}^{2}$ pairs of singers, each pair performing $r$ times, for a total of $r C_{6}^{2}=15 r$ performances. On the other hand, $m$ performances have $m \mathrm{C}_{4}^{2}=6 m$ pairs of singers performing together, thus,
$6 m=15 r, 2 m=5 r, 2|r, 5| m$.
If $m=5$, then $r=2$, meaning each pair performs together 2 times. 5 performances have a total of 20 appearances, with 6 people, and by the pigeonhole principle, there must be 1 person $A$ who appears at least 4 times. In these 4 performances, apart from $A$, there are 12 appearances, all by the other 5 people, and by the pigeonhole principle again, there must be 1 person $B(B \neq A)$ who appears at least 3 times. This leads to the pair $(A, B)$ appearing together at least 3 times in 5 performances, which is a contradiction.

If the minimum value of $m$ is 10, then $r=4$, meaning each pair performs together 4 times. 10 performances have a total of 40 appearances, and by the pigeonhole principle, there must be 1 person $C$ who appears at most 6 times. In the 6 performances where $C$ appears, apart from $C$, there are 18 appearances, all by the other 5 people, and by the pigeonhole principle again, there must be 1 person $D$ who appears at most 3 times. Thus, the pair $(C, D)$ appears together at most 3 times in 10 performances, which is a contradiction.

When $m=15$, number the 6 singers as $\{1,2,3,4,5,6\}$, and the 15 performances can be arranged as follows:
$\{1,2,3,4\},\{1,2,3,5\},\{1,2,3,6\},\{1,2,4,5\}$,
$\{1,2,4,6\},\{1,2,5,6\},\{1,3,4,5\},\{1,3,4,6\}$,
$\{1,3,5,6\},\{1,4,5,6\},\{2,3,4,5\},\{2,3,4,6\}$,
$\{2,3,5,6\},\{2,4,5,6\},\{3,4,5,6\}$.
Note that these 15 four-element subsets are exactly all the four-element subsets of the six-element set $\{1,2,3,4,5,6\}$, and the 6 elements are symmetric, with each pair appearing exactly 6 times, which meets the requirements of the problem.
In summary, the minimum value of $m$ is 15.
(2) $n=7$. At this time, 7 singers can form $\mathrm{C}_{7}^{2}=21$ pairs of singers, each pair performing $r$ times, for a total of $21 r$ performances. On the other hand, each performance has 4 people, and $m$ performances have $m \mathrm{C}_{4}^{2}=6 m$ pairs of singers performing together, so,
$$
6 m=21 r, 2 m=7 r, 2|r, 7| m \text {. }
$$

This indicates that at least $m=7$ performances need to be arranged.
When $m=7$, $r=2$. By the cyclic permutation method, the 7 performances can be arranged as follows:
$\{1,2,3,5\},\{2,3,4,6\},\{3,4,5,7\},\{4,5,6,1\}$,
$\{5,6,7,2\},\{6,7,1,3\},\{7,1,2,4\}$,

where each pair of four-element subsets has exactly 2 common elements, and each pair of $S=\{1,2, \ldots, 7\}$ belongs to exactly 2 of the above 7 four-element subsets, meaning each pair of 7 people appears together exactly 2 times.
(3) $n=9$. At this time, 9 singers can form $\mathrm{C}_{9}^{2}=36$ pairs of singers, each pair performing $r$ times, for a total of $36 r$ performances. Thus,
$$
6 m=36 r, m=6 r \text {. }
$$

If $m=6$, then $r=1$, meaning each pair appears together exactly 1 time. 6 performances have a total of 24 appearances, and by the pigeonhole principle, there must be 1 person $A$ who appears at least 3 times. In these 3 performances, apart from $A$, there are 9 appearances, all by the other 8 people, and by the pigeonhole principle again, there must be 1 person $B$ who appears at least 2 times. Thus, the pair $(A, B)$ appears together at least 2 times, which contradicts $r=1$.

If $m=12$, then $r=2$, meaning each pair appears together 2 times. 12 performances have a total of 48 appearances, and by the pigeonhole principle, there must be 1 person $C$ who appears at least 6 times. In these 6 performances, apart from $A$, the other 8 people have 18 appearances, and by the pigeonhole principle again, there must be 1 person $D$ who appears at least 3 times. Thus, the pair $(C, D)$ appears together at least 3 times, which contradicts $r=2$. This means the minimum value of $m$ is at least 18.

When $m=18$, $r=3$. By the cyclic permutation method, the 18 performances can be arranged as follows:
$\{1,2,3,5\},\{2,3,4,6\},\{3,4,5,7\}$,
$\{4,5,6,8\},\{5,6,7,9\},\{6,7,8,1\}$,
$\{7,8,9,2\},\{8,9,1,3\},\{9,1,2,4\}$;
$\{1,4,6,9\},\{2,5,7,1\},\{3,6,8,2\}$,
$\{4,7,9,3\},\{5,8,1,4\},\{6,9,2,5\}$,
$\{7,1,3,6\},\{8,2,4,7\},\{9,3,5,8\}$.
It is easy to verify that in these 18 performances, each pair appears together 3 times.
In summary, the minimum value of $m$ is 18.
(Li Chengzhang, Nankai University School of Mathematical Sciences, 300071)