Three. (25 points) A total of no more than 30 football teams from the East and West participate in the Super League. The East has 1 more team than the West. Any two teams play exactly one match. Each team earns 3 points for a win, 1 point for a draw, and 0 points for a loss. After the league ends, the statistics show that the ratio of wins to draws in matches between any two teams from the East, any two teams from the West, and between teams from the East and West is $2:1$. However, the total points of all teams from the East are 34 points more than those from the West. Find the number of teams from the East and the West that participated in the competition.
Three, let there be $x$ teams participating in the West, and in the matches between the East and the West, the East wins $y$ games, the West wins $z$ games, and there are $a$ draws.
According to the problem, we have
$$
(y+z): a=2: 1 \text {, }
$$

and $y+z+a=x(x+1)$.
Thus, $a=\frac{1}{3} x(x+1)$,
$$
z=\frac{2}{3} x(x+1)-y \text {. }
$$

Therefore, the number of wins in the internal matches of the West is
$$
\frac{2}{3} \times \frac{1}{2} x(x-1)=\frac{1}{3} x(x-1) \text {, }
$$

the number of draws is
$$
\frac{1}{3} \times \frac{1}{2} x(x-1)=\frac{1}{6} x(x-1) \text {; }
$$

the number of wins in the internal matches of the East is
$$
\frac{2}{3} \times \frac{1}{2} x(x+1)=\frac{1}{3} x(x+1) \text {, }
$$

the number of draws is
$$
\frac{1}{3} \times \frac{1}{2} x(x+1)=\frac{1}{6} x(x+1) \text {. }
$$

Since the total points of the East are 34 more than those of the West, we have
$$
\begin{aligned}
3 & \times \frac{1}{3} x(x+1)+2 \times \frac{1}{6} x(x+1)+3 y+a \\
= & 3 \times \frac{1}{3} x(x-1)+2 \times \frac{1}{6} x(x-1)+ \\
& 3 z+a+34 .
\end{aligned}
$$

Substituting equation (1) into the above equation and simplifying, we get
$$
9 y=3 x^{2}-x+51 \text {. }
$$

Clearly, $x$ is divisible by 3.
Let $x=3 k$ (where $k$ is a positive integer). Then
$$
y=3 k^{2}+6-\frac{k+1}{3} \text {. }
$$

Clearly, $\frac{k+1}{3}$ is a positive integer, denoted as
$\frac{k+1}{3}=t(t$ is a positive integer $)$.
Then $k=3 t-1 \Rightarrow x=9 t-3$.
By $x+(x+1) \leqslant 30 \Rightarrow x \leqslant 14 \Rightarrow t=1$.
Thus, $x=6, x+1=7$.
Therefore, there are 7 teams in the East and 6 teams in the West participating in the competition.