Three, (50 points) There are 12 football teams participating in a match, with each pair of teams playing one game. The winner gets 3 points, the loser gets 0 points, and in the case of a draw, both teams get 1 point. How many points does one team need to score at a minimum to ensure that no more than 6 teams score at least as many points as this team?
Three, assuming there are 7 teams with the same score as a certain team, then the maximum score for each match between these 8 teams is 3 points. Therefore, their maximum total score is $3 C_{8}^{2}=84$ points, and the score of each team cannot exceed $\left[\frac{84}{8}\right]=10$ points, so each team can win at most 3 matches.

Now, divide 12 teams into two groups, one with 8 teams of the same score, and the other with the remaining 4 teams. The 8 teams in the same group are arranged in a circle, where each team wins against the 3 teams behind it, draws with the 4th team, and loses to the other 3 teams. Thus, the maximum score for each of these 8 teams is $(3 \times 3+1)+3 \times 4=22$ points (i.e., each of the 8 teams in the same group wins against 3 teams, draws with 1 team, and loses to 3 teams, and wins against the 4 teams in the other group).

Therefore, when 8 teams have the same score (with 7 teams having a score no less than that team), each team can score a maximum of 22 points. Hence, when the proposition in the question holds, the score of that team is no less than 23 points.

Below is the proof: If there is a team that scores at least 23 points, then at most 6 teams can have a score no less than that team. Otherwise, if there are 7 teams with a score no less than 23 points, then the total score of these 8 teams is at least $8 \times 23=184$ points. On the other hand, dividing 12 teams into two groups, one with 8 teams and the other with 4 teams, the maximum total score for the 8 teams in the same group is $3 \mathrm{C}_{8}^{2}+3 \times 4 \times 8=180$ points. But $180 < 184$, which is a contradiction, so the proposition holds.