Suppose that there were $n$ players and assume $s_{1} \geq s_{2} \geq \cdots \geq s_{n}$ be the scores of each player. Because each player played exactly one game against each of the other players and each game worths 1 point, the total sum of scores is
$$s_{1}+s_{2}+\cdots+s_{n}=\binom{n}{2}=\frac{n(n-1)}{2}$$
Because each of the ten lowest scoring players earned half of his points against the other 9 of the 10, he earned the other half of his points against the $n-10$ highest scoring players. Therefore, the total sum of the scores earned in the games between the first $n-10$ players and the last 10 players is
$$10(n-10)=\frac{1}{2} s_{1}+\frac{1}{2} s_{2}+\cdots+\frac{1}{2} s_{n}=\frac{n(n-1)}{4} .$$
This implies that $n^{2}-41 n+400=0$, and thus $n=16$ or $25$.

Because each of the ten lowest scoring players earned half of his points against the other 9 of the 10, the total sum of the scores earned in the games between the last 10 players is
$$\frac{1}{2} s_{n-9}+\frac{1}{2} s_{n-8}+\cdots+\frac{1}{2} s_{n}=\binom{10}{2}=45 .$$
This gives an average of 9 points of score for each of the ten lowest scoring players. Therefore, each of the first $n-10$ players has earned at least 9 points.

Because each of the first $n-10$ has earned half of his score from the games against the $n-11$ of the $n-10$ players, we have
$$\frac{9}{2}(n-10) \leq \frac{1}{2} s_{1}+\frac{1}{2} s_{2}+\cdots+\frac{1}{2} s_{n-10}=\binom{n-10}{2}=\frac{(n-10)(n-11)}{2} .$$
We deduce that $n \geq 20$ and hence $n=25$.

Finally, we construct such a tournament with 25 players. In this tournament, there are 15 winners, each of them having 14 points of scores and 10 losers each of them having 9 points of score. Any game played between two winners or two losers results in a tie. For games played between winners and losers, each winner wins in 6 games, loses in 2 games, and 2 games result in a tie and conversely each loser wins 3 games, loses 9 games and 3 games result in a tie. This makes such a tournament possible.