Problem: In a competition, there are $a$ contestants
and $b$ judges, where $b \ge 3$ is an odd integer.
Each judge rates each contestant as either ``pass'' or ``fail''.
Suppose $k$ is a number such that for any two judges,
their ratings coincide for at most $k$ contestants.
Prove that
\[ \frac ka \ge \frac{b-1}{2b}. \]

Solution: This is a ``routine'' problem with global ideas.
We count pairs of coinciding ratings,
i.e.\ the number $N$ of tuples
\[(\{J_1, J_2\}, C) \]
of two distinct judges and a contestant
for which the judges gave the same rating.

On the one hand, if we count by the judges,
we have \[ N \le \binom b2 k \]
by he problem condition.

However, if we write $b=2m+1$ (so $m \coloneq \frac{b-1}{2}$), then each contestant $C$
contributes at least $\binom{m}{2} + \binom{m+1}{2} = m^2$ to $N$, and so
\[ N \ge a \cdot \left( \frac{b-1}{2} \right)^2 \]
Putting together the two estimates for $N$ yields the conclusion.