Finding the number of players in a tournament In a tournament, each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $1/2$ point if the game was a tie.
After the end of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten.)
What was the total number of players in the tournament?
 A: Given the phrasing of the problem we should assume that the tenth lowest score and the eleventh lowest score do not coincide.
Let $a_1\leq a_2\leq \dots \leq a_{10}$ be the $10$ lowest scores. Notice that exactly $\frac{a_1+a_2+\dots + a_{10}}{2}$ points were won in matches between lower scored players. On the other hand, since each game adds $1$ point to the total we have that exactly $\frac{10\cdot9}{2}=45$ points where won.
Therefore we have $a_1+a_2+\dots + a_{10}=90$
Notice that the number of points conceded by the $10$ lowest ranked players is $10(n-1)-90$. And so $10(n-1)-90=\frac{n(n-1)}{4}$, which is a quadratic with solutions $16$ and $25$.
$n=16$ is impossible, since we require the average number of points won by the $10$ lowest players to be $9$. While the total average is going to be $\frac{15}{2}<9$.
$n=25$ is possible.Make sure that all of the $10$ lowest players tie all of their games and all of the $15$ highest players tie all their games.
Then we just need to make sure that each lower player ties $9$ of his other games and loses $6$ and that each of the higher players ties $6$ games and wins $4$.
To do this we just need to select some edges of the complete bipartite graph $K_{10,15}$ so that every vertex on the small side has degree $6$ and every vertex on the big side has degree $4$. This is possible because of the Gale-Ryser theorem.
Although an explicit example can be found easily:Label the vertices in one side $1,2,\dots , 10$ and the others $1,2,\dots, 15$. Only connect vertex $x$ on the small side to the vertices $x-2,x-1,x,x+1,x+2\bmod 15$.
So only $n=25$ is possible.
