Yes, there always exists. Suppose the opposition that there exists a result of the tournament s.t. no $2$ players $A, B$ with $A$ not being weaker than $B$. In this result, no $2$ players can win in the same number of games playing white, otherwise the one with the more (could be the same) winning games playing black is not weaker than the other. Similarly, no $2$ players can win in the same number of games playing black. $\therefore$ the number of games that the player playing white wins is at least $0+1+\cdots+19$, and so is the number of games that the player playing black wins. $\Rightarrow$ there are at least $2(0+1+\cdots+19)>\binom{20}2$ games played, contradiction.