Five, in a chess tournament, there are an odd number of participants, and each participant plays one game against every other participant. The scoring system is as follows: 1 point for a win, 0.5 points for a draw, and 0 points for a loss. It is known that two of the participants together scored 8 points, and the average score of the others is an integer. How many participants are there in the tournament?

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Five, suppose there are $(n+2)$ players in total, except for 2 people who get 8 points, $n$ people on average get $k$ points each ($k$ is an integer).
$\because$ Each person plays one match with everyone else, and there are $(n+2)$ people,
$\therefore$ A total of $\frac{(n+1)(n+2)}{2}$ matches are played.
Since each match results in 1 point, we can set up the equation
$$
\frac{1}{2}(n+1)(n+2)=8+n k \text {, }
$$

Rearranging gives $n^{2}+(3-2 k) n-14=0$.
$\because$ The number of people is a positive integer, and $3-2 k$ is an integer,
$\therefore$ The number of people can only be $1, 2, 7, 14$.
Also, $\because$ the number of people is odd, $\therefore n=1$ or 7,
But when $n=1$, $k<0$ which is a contradiction, $\therefore n=7$.
At this time, $n+2=9$, i.e., a total of 9 people participate in the competition.