Problem 1550

1. Define Variables and Initial Conditions:
Let $n \geq 4$ be the number of players in the tournament. Each player $i$ has $w_i$ wins and $l_i$ losses. Since every player plays exactly one game with every other player, we have:
$$w_i + l_i = n - 1 \quad \text{for all } i.$$

2. Sum of Wins and Losses:
Since each game results in one win and one loss, the total number of wins is equal to the total number of losses:
$$\sum_{i=1}^n w_i = \sum_{i=1}^n l_i = S_1.$$

3. Sum of Squares of Wins and Losses:
The number of configurations where a player $A$ beats players $B$ and $C$ is given by:
$$\sum_{i=1}^n \binom{w_i}{2} = \sum_{i=1}^n \binom{l_i}{2}.$$
This implies:
$$\sum_{i=1}^n w_i^2 = \sum_{i=1}^n l_i^2 = S_2.$$

4. Sum of Cubes of Wins and Losses:
The number of configurations where a player $A$ beats players $B$, $C$, and $D$ is given by:
$$\sum_{i=1}^n \binom{w_i}{3} = \sum_{i=1}^n \binom{l_i}{3}.$$
This implies:
$$\sum_{i=1}^n w_i^3 \geq \sum_{i=1}^n l_i^3 = S_3.$$

5. **Expression for $\sum (w_i - l_i)^3$:**
Using the identity $w_i + l_i = n - 1 = m$, we can express $w_i - l_i$ as:
$$w_i - l_i = m - 2l_i.$$
Therefore:
$$\sum_{i=1}^n (w_i - l_i)^3 = \sum_{i=1}^n (m - 2l_i)^3.$$

6. Expand and Simplify the Cubic Expression:
Expanding $(m - 2l_i)^3$:
$$(m - 2l_i)^3 = m^3 - 6m^2 l_i + 12m l_i^2 - 8l_i^3.$$
Summing over all players:
$$\sum_{i=1}^n (m - 2l_i)^3 = \sum_{i=1}^n (m^3 - 6m^2 l_i + 12m l_i^2 - 8l_i^3).$$
This simplifies to:
$$\sum_{i=1}^n (m^3 - 6m^2 l_i + 12m l_i^2 - 8l_i^3) = nm^3 - 6m^2 S_1 + 12m S_2 - 8 S_3.$$

7. **Relate to $\sum (l_i - w_i)^3$:**
Similarly, we have:
$$\sum_{i=1}^n (l_i - w_i)^3 = \sum_{i=1}^n (- (w_i - l_i))^3 = -\sum_{i=1}^n (w_i - l_i)^3.$$
Therefore:
$$\Theta = \sum_{i=1}^n (w_i - l_i)^3 = -\Theta.$$

8. Conclusion:
Since $\Theta = -\Theta$, it follows that:
$$\Theta \geq -\Theta \implies \Theta \geq 0.$$

The final answer is $\boxed{ \sum_{i=1}^n (w_i - l_i)^3 \geq 0 }$.