9. A. 2 eighth-grade students and $m$ ninth-grade students participate in a single round-robin chess tournament, where each participant plays against every other participant exactly once. The scoring rule is: the winner of each match gets 3 points, the loser gets 0 points, and in the case of a draw, both players get 1 point. After the tournament, the total score of all students is 130 points, and the number of draws does not exceed half of the total number of matches. Then the value of $m$ is $\qquad$ .
9. A. 8 .

Let the number of draws be $a$, and the number of wins (losses) be $b$.
From the problem, we know
$$
2 a+3 b=130 \text {. }
$$

This gives $0 \leqslant b \leqslant 43$.
Also, $a+b=\frac{(m+1)(m+2)}{2}$
$$
\Rightarrow 2 a+2 b=(m+1)(m+2) \text {. }
$$

Thus, $0 \leqslant b=130-(m+1)(m+2) \leqslant 43$.
Therefore, $87 \leqslant(m+1)(m+2) \leqslant 130$.
From this, we get $m=8$ or 9 .
When $m=8$, $b=40, a=5$;
When $m=9$, $b=20, a=35, a>\frac{55}{2}$, which does not meet the problem's conditions.
Hence, $m=8$.