IMC 2014 Q21

Firstly suppose that any two knights X and Y win \(x\) and \(y\) bouts respectively and that \(x\) is at least as large as \(y\). The difference between their total scores would be the same as if X had won \(x - y\) bouts and Y had won none, since each of the separate totals would have been reduced by the same amount, namely 20\(y\). A similar procedure applies to losses. For example, if X won 3 and lost 6, while Y won 8 and lost 2, the difference between their total scores is the same as if X won 0 and lost 4, while Y won 5 and lost 0. In each case the difference is 32.
This argument shows that, in the case of the Black Knight, B, and the Red Knight, R, the smallest number of bouts will be achieved when one of B, R wins all his bouts and the other loses all his bouts. Also B has to score one more point than R. The possible scores for the knight who wins all his bouts are 20, 40, 60, 80, 100, 120, … . while the possible scores for the knight who loses all his bouts are 17, 34, 51, 68, 85, 102, 119, 136, … .
The first two numbers to differ by 1 are 119 and 120. Thus the Black Knight has a total of 120 corresponding to winning all of his 6 bouts and the Red Knight has a total of 119 corresponding to losing all of his 7 bouts.