Three, in a football round-robin tournament, the 1st to 4th places scored 11 points, 9 points, 7 points, and 5 points respectively, and the 5th place's score is different from the 4th place's score (the winner gets 2 points, the loser gets 0 points, and in a draw, both get 1 point). How many teams participated in the competition? And what is the total score of all the teams?
Three, suppose there are $k$ teams participating in the competition, a total of $C_{k}^{2}=\frac{1}{2} k(k-1)$ matches are played, so the total score is $2 C_{k}^{2}=k(k-1)$.
The top 4 teams scored $11+9+7+5=32$ points, the last $k-4$ teams can score at most $4(k-4)$ points, thus we have

Therefore, $k=7$.
The remaining 3 teams scored a total of $7 \times 6-32=10$ points. Let the scores of the last three teams be $m, n, p$, then
$$
\left\{\begin{array}{l}
4 \geqslant m \geqslant n \geqslant p \geqslant 0, \\
m+n+p=10 .
\end{array}\right.
$$

It is easy to see that (*) has two sets of positive integer solutions $(4,3,3),(4,4,2)$.
Thus, there are 7 teams participating in the competition, and the scores of the teams are $11,9,7,5,4,4,2$ or $11,9,7,5,4,3,3$.
Solution 2 The champion scored 11 points, at least 6 matches were played, so there are at least 7 teams participating. If there are 8 teams participating, then $C_{8}^{2}=28$ matches are played, a total of 56 points are accumulated, and the top four teams accumulated 32 points, so the last four teams accumulated 24 points. Since the 4th place accumulated 5 points, the last four teams can accumulate at most $4 \times 4=16$ points $<24$ points, which is a contradiction, thus there are only 7 teams participating. Then a total of $C_{7}^{2}=21$ matches are played, a total of 42 points are accumulated, the last three teams accumulated 10 points, the average score of the last three teams is $\frac{10}{3}=3.3$ points, so there must be one team with 4 points, and there must be one team with 3 points (or less than 3 points).
Since no team among the last three teams has 5 points, $\because$ no team has 1 point,
thus there are two possible score situations for the last two teams: $4,4,2$ or $4,3,3$.
The score situations of the teams are as follows: 11 points, 9 points, 7 points, 5 points, 4 points, 4 points, 2 points or 11 points, 9 points, 7 points, 5 points, 4 points, 3 points, 3 points.
Solution 3 The first place scored 11 points, at least 7 teams participated, if there are 8 teams participating, the total score is $2 C_{8}^{2}=56$ points.
The last four teams should have $56-32=24$ points, so at least one team has 6 points or more, which is impossible, so there are only 7 teams participating.

With 7 teams participating, the total score is $2 C_{7}^{2}=42$ points, the last three teams scored 10 points, and the scores are less than or equal to 4 points, there are two possibilities: 4,3,3; 4,4,2. (omitted)