for each game, someone wins and someone loses because there are no draws. if each participant won a different number of games, then each participant would have had to lose a different number of games.

I think there would be nothing to prove if it was stated in the problem that each participant played a game against every other participant. Nevertheless it is stated they played at most once. One may think of a generalisation (might be difficult - I don't know or there is an easy counterexample): If there were k different "scores" counting wins, were there necessarily exactly k different "scores" counting losses.

Isn't this still trivial? Since everyone played at most once, max number of wins is $15$, but since everyone won a different number, it would have to be all from $0,1,2,....,15$, so then its just obvious

$\color{blue} \boxed{\textbf{SOLUTION}}$ There are $16$ participants and everyone played at most one game against another participant, So one can win a maximum of $15$ games. All of the $16$ participants win different number of games between $0-15,$ but between $0-15$ there is exactly $16$ numbers. So, The number of winning games for $16$ participants are, $0,1,2,3,...15$ Now Consider the person who wins $i$ games, So he has lost $15-i$ games, So for every $i \in$ $(0,15)$ gives a different $15-i$ that is different number of losing games $\blacksquare$

kfas wrote: In a badminton tournament there were 16 participants. Each pair of participants played at most one game and there were no draws. After the tournament it turned out that each participant has won a different number of games. Prove that each participant has lost a different number of games. Thus, each $0$, $1$, $\dots$, $15$ wins is recorded. The $15$ wins thus $0$ losses, and rest has at least $1$. Simply induct.