169. In a chess tournament, two 7th-grade students and a certain number of 8th-grade students participated. Each participant played one game with every other participant. The two 7th-graders together scored 8 points, and all the 8th-graders scored the same number of points (in the tournament, each participant earns 1 point for a win and $1 / 2$ point for a draw). How many 8th-graders participated in the tournament?
169. Let $n$ be the number of eighth-graders, and $m$ be the number of points earned by each of them. In this case, the number of points scored by all participants in the tournament is $m n + 8$. This number is equal to the number of games played. Since the number of participants in the tournament is $n + 2$ and each played once with each of the $n + 1$ others, they played a total of $\frac{(n+2)(n+1)}{2}$ games (in the product $(n+2)(n+1)$, each game is counted twice). Thus, we have:

$$
m n + 8 = \frac{(n+2)(n+1)}{2}
$$

or, after some transformations,

$$
n(n + 3 - 2m) = 14
$$

Here, $n$ is an integer; the number in parentheses is also an integer (since $m$ is either an integer or a fraction with a denominator of 2).

Since $n$ is a divisor of 14, $n$ can be one of the numbers $1, 2, 7, 14$. The values $n = 1$ and $n = 2$ should be discarded, as in these cases, the total number of participants does not exceed 4, and two seventh-graders could not have scored a total of 8 points.

This leaves $n = 7$ and $n = 14$.

If $n = 7$, then $7(7 + 3 - 2m) = 14; m = 4$.

If $n = 14$, then $14(14 + 3 - 2m) = 14; m = 8$.