Consider the related problem: For each $n\in N$ which is the minimum number $f(n)$ such that there exists a tournament where someone has at some point $n$ wins? For example $f(1)=2, f(2)=3, f(3)=5$, etc. Looks like Fibonacci... Indeed, let's prove $f(n+1)=f(n)+f(n-1)$. To have someone with $n+1$ wins, a player with $n$ wins must beat another player with $n-1$ wins. Since there must be (at least) $f(n)$ players for someone to have $n$ wins, the same for $n-1$, and all those players are distinct (because losers drop out), we have $f(n+1)=f(n)+f(n-1)$. Since $f(1)=2, f(2)=3$, we have in general $f(n)=F_{n+2}$ where $F_n$ is the nth Fibonacci number. Since $F_{10}=55$ we have $n=8$; for $n>8$ you need more than 55 people.