# What is the number of the players in the tournament? (Logic puzzle)

I have been struggling with the following problem I would appreciate some hints : In a tournament every player can play against each of the other players exactly once.In each game, the winner gains 1 point and the loser 0 point and if they tie, they each get 1/2 point.At the end of the tournament , each player has exactly half of her/his score against the 10 players with the least scores .(And each of the players with the least scores has gained exactly half of their score against the other 9 players among them) What is the number of the players in the tournament?

Let $$L$$ be the set of the $$10$$ losing player and $$W$$ be the set of the remaining players. Suppose there are $$n$$ people in the group $$W$$. Let $$P_{LL}, P_{LW}, P_{WL}, P_{WW}$$ be the points $$L$$ got from $$L$$, $$L$$ got from $$W$$, $$W$$ got from $$L$$ and $$W$$ got from $$W$$, respectively.
Then we know that $$P_{LL} = P_{LW}$$ and $$P_{WL} = P_{WW}$$. Also, $$P_{LL}$$ is $$\binom{10}2$$ since on each game played among $$L$$, the players in $$L$$ got $$1$$ point in total, no matter there is a tie or not. Similarly, $$P_{WW} = \binom n2$$.
We also know that there are $$10n$$ games played between $$L$$ and $$W$$, so $$P_{LW}+P_{WL} = 10n$$.
All these conditions will give you a quadratic equation in $$n$$. Solve it. You will get two integral solutions. One of them will be extraneous, since in that extraneous solution. It turns out that the losing players will indeed win more games than the winning players, which is impossible. So you get one unique solution.
Let $$A$$ be the $$n \times n$$ matrix, where $$A_{ij}$$ is the score of player $$i$$ against player $$j$$ if $$i \neq j$$, and $$0$$ if $$i = j$$. Let $$\mathbf{1}$$ be the vector of $$n$$ ones. What can you say about the number $$\mathbf{1}^T A \mathbf{1}$$? And how can you model the other conditions?