Solution:

Denote by $x$ the number of African teams. Then the number of European teams equals $x+9$. The African teams played each other $\frac{(x-1)x}{2}$ games and therefore the points won by them are $\frac{(x-1)x}{2} + k$, where $k$ is the number of wins over European teams.

Further, the points won by the Europeans are $\frac{(x+8)(x+9)}{2} + x(x+9) - k$. Thus,
$$9\left(\frac{(x-1)x}{2} + k\right) = \frac{(x+8)(x+9)}{2} + x(x+9) - k$$
and so $3x^2 - 22x + 10k - 36 = 0$. Since $x$ is a positive integer, we have that $121 - 3(10k - 36) = 229 - 30k$ is a perfect square. Then $k \leq 7$ and a direct verification shows that we obtain perfect squares only for $k=2$ and $k=6$. For $k=2$ we have $x=8$ and therefore the best African team could have at most $7+2=9$ points.

For $k=6$ we get $x=6$ and therefore there are 6 African and 15 European teams. In this case the best African team has at most $5+6=11$ points, which happens if it wins over all other African teams and 6 European teams (the other African teams lost their games against all European teams). Finally, the answer is $11$.