Source Serbian Mathematical Olympiad 2015
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Assume the contrary. For each $i = 1,..., 100$ choose a candidate $A_i$ who was voted for exactly $i$ times.
The number of judges who gave both their votes for candidates in the set
$A = \\{A_{34}, A_{35},..., A_{100}\\}$
does not exceed the number of pairs actor-actress in $A$, and the number of such pairs is at most $33 \cdot 34 = 1122$.
On the other hand, of the $2 \cdot 3366 = 6732$ votes, exactly
$34 + 35 +···+ 100 = 4489$
were given to the candidates in $A$.
Therefore at most $6732 − 4489 = 2243$ judges could have given a vote to a candidate not in $A$.
Thus, there were at most $1122 + 2243 = 3365$ judges, a contradiction.