We label the players as $A_{1}, A_{2}, A_{3},\cdots, A_{100}$ depending on the leaderboard results($A_{1}$ got the most points and $A_{100}$ has the least points). Now lets say that we look at an example for $k$. If we look at the group ${A_{k+1}, A_{k+2},\cdots, A_{100}}$: They have played exactly $\frac{(100-k)(99-k)}{2}$ mathes. That means that the sum of the results of these people is at least $(100-k) + \frac{(100-k)(99-k)}{2}$, because $A_{k}$ lost to all people of this group. Now we know that $A_{k}$ has more points than any of the people of this group, so we derive the inequality: $(k-1)(100-k) > (100-k)+\frac{(100-k)(99-k)}{2}$. We know its true, because $A_{k}$ has $k-1$ points and we know that $A_{k}$ has more points than any player of ${A_{k+1}, A_{k+2},\cdots, A_{100}}$, so we multiply by $100-k$. I really hope the inequality deriavtion is clear enough. Now we solve the inequality: $(k-1)(100-k) > (100-k)+\frac{(100-k)(99-k)}{2} \implies (k-1) > 1 +frac{99-k}{2} \implies 2k-2 > 2 + 99 - k \implies 3k > 103$, so $k>34$. We get a setup for $k=35$: We look at the graph with vertices $A_{36},\cdots , A_{100}$. By induction we can see that we can construct a directed graph( the direction decides the winner of the match) with $2n+1$ vertices and n directed edges coming out of each vertex. We construct the cycles $A_{36},A_{37},\cdots,A_{100}; A_{36},A_{38},A_{40}\cdots; A_{36}, A_{39}, A_{42}$and so on. If we have a natural number $r>1$, for which $r\vert 65$, then we have more than one (exactly $\frac{65}{r}$) cycles $A_{36},A_{36+r},A_{36+2r},\cdots$, but also $A_{37},A_{37+r},A_{37+2r}\cdots$ and so on to the cycle $A_{35},A_{35+r},A_{35+2r},A_{35+3r}$. Now every cycle is directed (consists of directed edges) and either $(A_{i}, A_{j})$ or $(A_{j}, A_{i})$ is an edge. Inductively we have that every vertex has $32$ edges coming out of it. As $A_{35}$ has lost to all $A_{36},\cdots A_{100}$. So every $A_{36},\cdots A_{100}$ has $32+1$ points and $A_{35}$ has $34$ points. Now if we set that every $A_{i}$ for any $0<i<35$ has won every match to all $A_{36},\cdots A_{100}$, so every $A_{1},A_{2},\cdots,A_{34}$ has won at least $6$5 matches (all the matches between two of $A_{1},A_{2},\cdots,A_{34}$ we set the winner at random: for example only we set $A_{i}$ to have beaten $A_{j}$ if $i<j$ for $0<i<j<35$), so $A_{35}$ is in the $35$-th spot and we are done!