Problem 1387

1. Each team plays with every other team exactly once, so the total number of games each team plays is $n - 1$. Therefore, for each team $i$, we have:
$$x_i + y_i = n - 1$$
where $x_i$ is the number of wins and $y_i$ is the number of losses for team $i$.

2. We need to prove that:
$$\sum_{i=1}^{n} x_i^2 = \sum_{i=1}^{n} y_i^2$$

3. Start by considering the expression:
$$\sum_{i=1}^{n} (x_i^2 - y_i^2)$$

4. Using the identity $a^2 - b^2 = (a + b)(a - b)$, we can rewrite the sum as:
$$\sum_{i=1}^{n} (x_i^2 - y_i^2) = \sum_{i=1}^{n} (x_i + y_i)(x_i - y_i)$$

5. Since $x_i + y_i = n - 1$ for all $i$, we can factor out $n - 1$ from the sum:
$$\sum_{i=1}^{n} (x_i + y_i)(x_i - y_i) = (n - 1) \sum_{i=1}^{n} (x_i - y_i)$$

6. Next, observe that the total number of wins in the tournament is equal to the total number of losses because each game results in one win and one loss. Therefore:
$$\sum_{i=1}^{n} x_i = \sum_{i=1}^{n} y_i$$

7. This implies:
$$\sum_{i=1}^{n} (x_i - y_i) = \sum_{i=1}^{n} x_i - \sum_{i=1}^{n} y_i = 0$$

8. Substituting this back into our earlier expression, we get:
$$(n - 1) \sum_{i=1}^{n} (x_i - y_i) = (n - 1) \cdot 0 = 0$$

9. Therefore:
$$\sum_{i=1}^{n} (x_i^2 - y_i^2) = 0$$

10. This implies:
$$\sum_{i=1}^{n} x_i^2 = \sum_{i=1}^{n} y_i^2$$

$\blacksquare$