44. There are $n$ players participating in a round-robin chess tournament, meaning all players play against each other exactly once. Each match, the winner gets 3 points, a draw results in 1 point for each player, and the loser gets 0 points. The total points of all players at the end is 120. Then the maximum number of participants is $\qquad$ people.
Answer: 11
Solution: Let the number of draws be $a$ and the number of win-loss games be $b$, then we have
$$
\begin{array}{l}
2 a+3 b=120 \\
0 \leq b \leq 40
\end{array}
$$

Given $a \geq 0$, and $a$ is an integer, let the total number of participants be $n$, then
$$
a+b=\frac{n(n-1)}{2} \text {, }
$$

which means
$$
2 a+2 b=n(n-1)
$$

From (1) and (2), we get
$$
b=120-n(n-1),
$$

thus
$$
0 \leq 120-n(n-1) \leq 40 \text {, }
$$

which simplifies to
$$
80 \leq n(n-1) \leq 120 \text {, }
$$

The maximum value of $n$ that meets the requirement is 11, so the maximum number of participants is 11.