Note that , if a match was won by one of the two teams then the sum of the points two teams get is $3$ and for draw it is $2$. Assume that $x$ matches were won and there were draws in $y$ matches.The total number of matches played is $ \binom{20}{2}=190$ Then $3x+2y=554$ $x+y=190$ Solving the equations we get $x=174$ and $y=16$. So, in $16$ matches there were draws. Now we need to find the minimum number of teams needed to play $16$ matches. As $n$ teams can play maximum $ \binom{n}{2}$ matches ,we need to find the minimum $n$ such that $\binom{n}{2} \geq 16$ As $\binom{6}{2}=15$ and $\binom{7}{2}=21$. The value of $n$ is $7$. So, there exists $7$ teams having at least one draw.