One, (20 points) A competitive event involves several teams from two cities, A and B. It is known that city B has 8 more teams than city A, and any two teams play exactly one match. The event rules state: the winner gets 1 point, the loser gets 0 points, and there are no ties. In the end, the total score of all teams from city B is 4 points more than the total score of all teams from city A. Find the minimum score of the best team from city A.
Let city A have $x$ teams participating in the competition. Then city B has $x+8$ teams. When the teams from cities A and B compete against each other, city A scores a total of $y$ points, and city B scores a total of $x(x+8)-y$ points.
From the problem, we have
$$
\begin{array}{l}
\frac{1}{2} x(x-1)+y+4 \\
=\frac{1}{2}[(x+8)-1](x+8)+[x(x+8)-y] \\
\Rightarrow y=\frac{1}{2} x^{2}+8 x+12 .
\end{array}
$$

Since $y$ is an integer, $x$ must be even.
$$
\begin{array}{l}
\text { From } x(x+8)-y \geqslant 0 \\
\Rightarrow x(x+8)-\left(\frac{1}{2} x^{2}+8 x+12\right) \geqslant 0 \\
\Rightarrow x \geqslant \sqrt{24}>4 .
\end{array}
$$

Since $x$ is even, the minimum value of $x$ is 6. Clearly, $y \geqslant x+8$.
Thus, the highest score of the best team in city A is
$$
(x-1)+(x+8)=2 x+7 \text { (points). }
$$

Therefore, when $x=6$, the minimum value of $2 x+7$ is 19. Hence, the minimum value of the highest score of the best team in city A is 19.