In a mathematics competition, 300 participants take part. After the competition, some participants play chess. Each pair of participants plays at most one game against each other. There are no three participants at this competition who all play against each other. Determine the maximum $n$ for which it is possible to satisfy the following conditions simultaneously: each participant plays at most $n$ chess games, and for every $m$ with $1 \leq m \leq n$ there is a participant who plays exactly $m$ chess games.
We show that $n$ is at most 200. First, we give an example where $n=200$. Consider players $A_{1}, A_{2}, \ldots, A_{200}$ and $B_{1}, B_{2}, \ldots, B_{100}$. This makes a total of 300 players. Let $B_{i}$ play chess against the players $A_{j}$ with $1 \leq j \leq i+100$, while no other participants play against each other. Then $B_{i}$ plays exactly $100+i$ other participants, so for $101 \leq m \leq 200$ there is a participant who plays exactly $m$ games. Furthermore, $A_{j}$ for $j>100$ plays against the participants $B_{i}$ with $i \geq j-100$; there are $100-(j-101)=201-j$ of them. Since $j$ can vary from 101 to 200, this number varies from 100 to 1. Thus, for $1 \leq m \leq 100$ there is also a participant who plays exactly $m$ games. Finally, $A_{j}$ for $j \leq 100$ plays against all $B_{i}$; there are 100 of them. Therefore, there are no participants who play more than 200 games. We see that this example meets the requirements for $n=200$.
Now we prove that $n>200$ is not possible. We do this by contradiction, so assume that $n>200$. Then there is at least one participant $A$ who plays exactly 201 games, against players $B_{1}, B_{2}, \ldots, B_{201}$. There are a total of 300 participants, so apart from them and $A$, there are still $300-1-201=98$ other participants, which we call $C_{1}, \ldots, C_{98}$. If a participant $B_{i}$ plays against another participant $B_{j}$, they form a trio with $A$ who play three games among themselves; this is not allowed. Therefore, $B_{i}$ does not play against another participant $B_{j}$. So he plays at most $1+98=99$ games. This means that the participants who play exactly $m$ other participants with $100 \leq m \leq 200$ must all be among the participants $C_{i}$. However, there are only 98 different people, a contradiction. We conclude that $n>200$ is not possible and that $n=200$ is the maximum.