The Reds, the Grays, the Blues, and the Blacks have a round-robin tournament. Each team plays each other team once, for a total of six games.
The Blacks won more games than the Blues. The Grays lost more games than the Blues. The Reds tied the Blacks. (This was the only tie in the tournament.) Who won the game between the Reds and the Blues?
The Reds beat the Blues
As the Blacks tied a game, they have at most two wins. The Blues have at most one win. As the Grays have at most three losses, the Blues have at most two losses. The Blues did not tie a game, so they finished with one win, two losses. The Blacks have two wins and a tie and the Grays have three losses. To balance out the wins and losses the Reds have two wins and a tie. Since the Reds tied the Blacks, they beat the Blues.
The Blues and Greys never drew, yet the Blues had fewer losses than the Greys, which means that they also had more wins. Since the Blacks drew against the Reds, they have at most two wins, and they have more wins than the Blues, who have at least one.
Blacks have two wins and a draw, Blues have one win and two losses, and Greys have three losses. The solution can be derived several different ways:
1) There must be five wins in total, the remaining two of which could only be won by the Reds. One of these is against the Blues.
2) The Greys lost every match, including the one against Blues, so one of the Blues' two losses must be against the Reds.
Blacks ties exactly one game. So they could have won at most 2 games.
Case 1 Let's say that the Blacks won 2 games.
Case 1a) Blues won 1 game and lost 2. In this case, Grays must have lost all their games. This means that Reds won 2 games. They must have won these 2 games against Blues and Grays.
Case 1b) Blues won 0 game and lost 3. This isn't possible because the Grays lost more games than the Blues.
Case 2 Blacks won 1 game. In this case, Blues won 0 and lost 3 games. This isn't possible because Grays have to have lost more games than the Blues.
So, only case 1a is possible. Thus, the answer is that Reds defeated the Blues in their game.
Aside from the accepted answer, this puzzle has a second solution. Blue could have beaten Red in the following case:
Red and Black both won 2 games, Blue won 1 and drew 1, and grey drew 1. This happens if Black loses to Red, but wins from the other two, while Blue and Grey drew, and Grey lost all other games.
We end up with the following scores (I assume drawing gives half a point, but it does not matter):
