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\:\large\frac{n(n-1)}{2}$$-2n+9=84$

$\Rightarrow\:n^2-5n-150=0$

$\Rightarrow\:n=15\:or\:-10 \:(not\: possible)$

$\Rightarrow\:n=15$