1985 AIME #14

Notice that in every match number of points earned is exactly $1$.Now let $n$ be the number of non-lowest $10$ contestants.Among themselves they played $\binom {n}{2}$ matches and earned that many points.Now lowest $10$ among them earned $45$ points playing with each other so they earned $45$ points playing against the non-lowest $10$ players.Now that means that the non-lowest $10$ players against the lowest $10$ earned $10n-45$.So we obtain $10n-45=\frac{n(n-1)}{2}$.By solving it we get $n=15$ or $n=6$.It is trivial that $n$ can't be $6$ so the answer is $25$.

That is wrong. You just said "we get n=15 or n=6... so the answer is 25". Here is a correct solution. Let the points earned by each player be s1,...,sn in increasing order. In each game the sum of the players points is 1, so s1+...+sn= total # games=(n choose 2). Considering the ten worst players we similarly get (1/2)(s1+..+s10)=total # games among 10 worst players=(10 choose 2), so s1+...+s10=90. Among everyone else, half the points are earned against the 10 worst players, so the other half of the points are earned amongst themselves. That is, (1/2)(s11+..sn)=((n-10)choose 2), so s11+...+sn=(n-10)(n-11). Thus s1+...+sn=90+(n-10)(n-11)= n choose 2=(1/2)n(n-1) which you solve to get n=16,25. Since s1+...s10=90, s10 is at least 9, so s11+..s16 is at least 6*0=54 so s1+....+s16 is at least 90+54=144. But s1+..s16= (16 choose 2)=120, a contradiction. Thus n=25.

Solution. There are $\binom{n}{2}$ points in total. Note that the condition tells us that half of the points of the total, $\frac{1}{2}\cdot \binom{n}{2}$ is from the games between the $10$ and the $n-10$ players, so $$\frac{1}{2}\cdot \binom{n}{2} = (n-10)10$$$$n = 16 \text{ or } n = 25$$If $n$ is $16$, it's easy to calculate that the average of the $6$ biggest players is $\frac{30}{6}=5$, while the average of the $10$ smallest is $\frac{120-30}{10} = 9$, a contradiction. So $n=\boxed{25}$.