Let the no. of participants be $n$.
If no participant fell ill, then the no. of games played = $^nC_2$
Each player will play $n-1$ games in the tournament.
These n-1 games include 1 game which both the player played with each other.
This one game is counted with both the players.
The two participants played 3 games each and fell ill.
$\therefore$ The no. of games played = $^nC_2-[(n-1)+(n-1)-1]+6=84$ (given)
$\Rightarrow\:n=15\:or\:-10 \:(not\: possible)$