9,10

In a chess tournament, 12 people participated. After the tournament, each participant made 12 lists. The first list includes only themselves, the second list includes them and those they defeated, the third list includes everyone from the second list and those they defeated, and so on. The 12th list includes everyone from the 11th list and those they defeated. It is known that for any participant in the tournament, the 12th list includes a person who was not in the 11th list. How many draws were played in the tournament?
Answer: 54. If the $(k+1)$-th list is the same as the $k$-th, then the lists numbered $k+2, \ldots, 11, 12$ will also be exactly the same. But according to the condition, the 11-th list and the 12-th are different. Therefore, each participant's $k$-th list contains exactly $k$ people. In particular, the 2-nd list contains exactly two people. This means that each participant won exactly one game. Therefore, the number of draws is $\frac{12 \cdot 11}{2}-12=54$.