Problem 830

Let the number of participants from the second year be $n$, so together with the first year, there are $n+2$ participants, hence the number of matches played is $C_{n+2}^{2}$, and the total score of all participants is $C_{n+2}^{2}$. Among these, 8 points were scored by the first-year students, and the remaining $C_{n+2}^{2}-8$ points were evenly distributed among the second-year students. Therefore, according to the problem, we have
$$\frac{C_{n+2}^{2}-8}{n}=\text { a non-negative integer }$$

which means
$$\frac{n}{2}+\frac{3}{2}-\frac{7}{n}=\text { a non-negative integer }$$

Thus, if $n$ is odd, then $7 / n$ must be an integer, which implies $n=7$ or 1. However, $n=1$ contradicts the fact that the two first-year students scored a total of 8 points, so $n=7$; if $n$ is even, then $3 / 2-7 / n$ must be an integer, which implies $7 / n$ must be a reduced fraction with a denominator of 2, hence $n=14$.