Time to train!

Let the number of events be $n$ . Since three houses competed and three places were awarded in every event, all the points awarded in $n$ events were received by the three houses. Therefore, we have $n(x+y+z) = 22+9+9 = 40$ . Since the minimum of $s=x+y+z$ is $3+2+1=6$ , $n < 7$ . As both $n$ and $s$ are positive integers, $(n,s)$ can only be $(1,40)$ , $(2,20)$ , $(4,10)$ and $(5,8)$ .

• If $n=1$ , then one house would have scored $x$ , one $y$ and the last $z$ and Slytherin and Ravenclaw would not score the same. Therefore, $n \ne 1$ .
• If $n=2$ , then Glyffindor would have scored $x+y$ , Slytherin, who won the Quidditch event, $x+z$ , and Ravenclaw, $y+z$ . But $x+z \ne y+z$ , so $n \ne 2$ .
• If $n=4$ , then $s=10$ . The possible $(x,y,z) = (7,2,1), (6,3,1),(5,4,1), (5,3,2)$ . Since Slytherin won an $x$ , the only possible overall score was $x+z+z+z = 6+1+1+1 = 9$ . But no other combination from the set $(x,y,z)=(6,3,1)$ could yield 9 for Ravenclaw. So $n\ne 4$ .
• Then $n$ must be 5 and $s=8$ . The possible $(x,y,z) = (5,2,1), (4,3,1)$ . The only possible Slytherin's overall score is $x+z+z+z+z = 5+1+1+1+1=9$ . Ravenclaw's overall score is $y+y+y+y+z=2+2+2+2+1=9$ and Gryffindor's, $x+x+x+x+y=5+5+5+5+2=22$ .

Overall placing of the three houses

$\begin{array} {cccccc} \text{Place} & \text{Event 1} & \text{Event 2} & \text{Event 3} & \text{OWL} & \text{Quidditch} \\ \hline \text{1st} & \color{#D61F06} Gryffindor & \color{#D61F06} Gryffindor & \color{#D61F06} Gryffindor & \color{#D61F06} Gryffindor & Slytherin \\ \text{2nd} & \color{#3D99F6}Ravenclaw & \color{#3D99F6}Ravenclaw & \color{#3D99F6}Ravenclaw & \color{#3D99F6}Ravenclaw & \color{#D61F06} Gryffindor \\ \text{3rd} & Slytherin & Slytherin & Slytherin & Slytherin & \color{#3D99F6}Ravenclaw \end{array}$

Therefore, $\color{#3D99F6}\boxed{Ravenclaw}$ finished second in OWL.