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 $\tfrac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned in games against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of his points against the other nine of the ten). What was the total number of players in the tournament?
1. **Total Points Calculation:**
   In a tournament with \( n \) players, each player plays exactly one game against each of the other players. Therefore, the total number of games played is given by the combination formula:
   \[
   \binom{n}{2} = \frac{n(n-1)}{2}
   \]
   Each game results in 1 point being distributed (either 1 point to the winner or 0.5 points to each player in case of a tie). Thus, the total number of points distributed in the tournament is:
   \[
   \frac{n(n-1)}{2}
   \]

2. **Points Earned by the Ten Lowest Scoring Players:**
   Let the ten players with the least number of points be denoted as \( P_1, P_2, \ldots, P_{10} \). According to the problem, each of these ten players earned half of their points in games against the other nine of the ten. Therefore, the total points earned by these ten players in games among themselves is:
   \[
   \binom{10}{2} = \frac{10 \cdot 9}{2} = 45
   \]
   Since each game among these ten players results in 1 point being distributed, the total points distributed among these ten players in their games against each other is 45.

3. **Points Earned by the Ten Lowest Scoring Players Against the Other \( n-10 \) Players:**
   Let the total points earned by these ten players be \( T \). According to the problem, half of these points were earned in games against the other \( n-10 \) players. Therefore, the points earned by these ten players in games against the other \( n-10 \) players is:
   \[
   \frac{T}{2}
   \]
   Since the total points earned by these ten players is \( 2 \times 45 = 90 \), we have:
   \[
   T = 90
   \]
   Therefore, the points earned by these ten players in games against the other \( n-10 \) players is:
   \[
   \frac{90}{2} = 45
   \]

4. **Total Points Earned by the Other \( n-10 \) Players:**
   The total points earned by the other \( n-10 \) players is:
   \[
   \frac{n(n-1)}{2} - 90
   \]
   Since half of the points earned by each of these \( n-10 \) players were earned in games against the ten lowest scoring players, the total points earned by these \( n-10 \) players in games against the ten lowest scoring players is:
   \[
   45
   \]

5. **Equation Setup:**
   The total points distributed in the tournament is:
   \[
   \frac{n(n-1)}{2}
   \]
   The points distributed in games between the ten lowest scoring players and the other \( n-10 \) players is:
   \[
   45 + 45 = 90
   \]
   Therefore, we have:
   \[
   \frac{1}{2} \cdot \frac{n(n-1)}{2} = 90
   \]
   Simplifying, we get:
   \[
   \frac{n(n-1)}{4} = 90
   \]
   \[
   n(n-1) = 360
   \]

6. **Solving the Quadratic Equation:**
   Solving the quadratic equation \( n(n-1) = 360 \):
   \[
   n^2 - n - 360 = 0
   \]
   Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -1 \), and \( c = -360 \):
   \[
   n = \frac{1 \pm \sqrt{1 + 1440}}{2}
   \]
   \[
   n = \frac{1 \pm \sqrt{1441}}{2}
   \]
   Since \( n \) must be a positive integer, we take the positive root:
   \[
   n = \frac{1 + 37}{2} = 19
   \]

The final answer is \( \boxed{19} \).