Three, 18 football teams are participating in a single round-robin tournament, meaning each round the 18 teams are divided into 9 groups, with each group's two teams playing one match. In the next round, the teams are regrouped to play, for a total of 17 rounds, ensuring that each team plays one match against each of the other 17 teams. After $n$ rounds of the tournament according to any feasible schedule, there always exist 4 teams that have played a total of only 1 match among themselves. Find the maximum possible value of $n$.
(Li Chengzhang, contributor)
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3. Consider the following competition program:
1.(1.2)(3.4)(5.6)(7.8)(9,18)
    (10,11)(12,13)(14,15)(16,17)
2.(1,3)(2.4)(5,7)(6.9)(8,17)
    (10,12)(11,13)(14,16)(15.18)
3.(1.4)(2.5)(3.6)(8.9)(7.16)
    (10,13)(11,14)(12,15)(17,18)
4.(1,5)(2.7)(3.8)(4,9)(6.15)
    (10.14)(11,16)(12.17)(13.18)
5.(1,6)(2,8)(3,9)(4,7)(5,14)
    (10,15)(11,17)(12,18)(13.16)
6.(1,7)(2.9)(3.5)(6,8)(4,13)
    (10,16)(11,18)(12,14)(15,17)
7.(1.8)(2.6)(4.5)(7.9)(3.12)
    (10.17)(11,15)(13,14)(16,18)
8. (1,9)(3,7)(4,6)(5,8)(2,11)
    (10,18)(12,16)(13.15)(14.17)
9.(2.3)(4.8)(5.9)(6.7)(1.10)
    (11,12)(13,17)(14,18)(15,16)
10.(1,11)(2.12)(3,13)(4,14)(5,15)
        (6,16)(7,17)(8,18)(9,10)
11.(1.12)(2,13)(3,14)(4,15)(5,16)
        (6,17)(7,18)(8,10)(9,11)
12.(1,13)(2.14)(3,15)(4,16)(5,17)
        (6,18)(7.10)(8.11)(9.12)
.....
17.(1,18)(2,10)(3,11)(4,12)(5,13)
    (6,14)(7,15)(8,16)(9,17)
The first 9 teams are called Group A, and the last 9 teams are called Group B. It is easy to see that after 9 rounds, any two teams in the same group have already played against each other. Therefore, any 4 teams have already played at least two matches, which of course does not meet the requirements of the problem.

If the above program is reversed and the matches are played in the new order, then after 8 rounds, any two teams in the same group have not played against each other. Each team has played 1 match against 8 of the 9 teams in the other group. At this point, any 4 teams in the same group have not played against each other, and any 4 teams not all in the same group have played at least two matches. Of course, none of these meet the requirements of the problem.
The above discussion shows that the maximum possible value of \( n \) is no more than 7.
Assume that 7 rounds of matches have been played and no 4 teams meet the requirements of the problem.

Select two teams \( A_1 \) and \( A_2 \) that have played one match. Thus, each team has played against 6 other teams. The two teams have played against at most 12 other teams. Therefore, there are at least 4 teams \( B_1, B_2, B_3, B_4 \) that have not played against \( A_1 \) and \( A_2 \). By the assumption of contradiction, \( B_1, B_2, B_3, \) and \( B_4 \) have all played against each other.
Teams \( B_1 \) and \( B_2 \) have each played 3 matches among the 4 teams \( \{B_1, B_2, B_3, B_4\} \). Thus, each team has played 1 match against 4 of the other 14 teams. This leads to at least 6 teams \( C_1, C_2, \cdots, C_6 \) that have not played against \( B_1 \) and \( B_2 \). By the assumption of contradiction, \( C_1, C_2, \cdots, C_6 \) have all played against each other.
Teams \( C_1 \) and \( C_2 \) have each played 5 matches among the 6 teams \( \{C_1, C_2, \cdots, C_6\} \). Thus, each team has played 1 match against 2 of the other 12 teams. Therefore, there are at least 8 teams \( D_1, D_2, \cdots, D_8 \) that have not played against \( C_1 \) and \( C_2 \). By the assumption of contradiction, these 8 teams have all played against each other. This way, \( D_1 \) and \( D_2 \) have not played against 10 other teams. Since only 7 rounds have been played, there must be two teams \( E_1 \) and \( E_2 \) among the 10 other teams that have not played against each other. Thus, \( D_1, D_2, E_1, \) and \( E_2 \) have played a total of 1 match, which contradicts the assumption of contradiction.
In summary, the maximum number of rounds that can be played is 7.
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