Candidates in an exam 443 candidates enter the exam hall. There are 20 rows of seats I'm the hall. Each row has 25 seats. At least how many rows have an equal number of candidates.
My attempt
Seat 25 in the first row  24 in the second and so on...but I cannot find the worst case scenario.
 A: Let $a_1,a_2\dots a_{20}$ be non-negative integers such that $a_1+a_2+\dots a_{20}=443$. Denote by $s$ the size of the largest subset of the $a$'s which all have the same value. I think we want to minimize this. 
We shall prove it is three. To show it is more than two we notice that with $s=2$ the maximum sum would be achieved with
$a_1=a_{11}=16,a_2=a_{12}=17\dots a_3=a_{3}=18 \dots a_{10}=a_{20}=25$
Then the sum of the $a$'s would be $2\frac{10(41)}{2}=410$.
We now prove with $s=3$ it is obtainable, to see this we take the numbers from $20$ to $25$ three times (these are $6\times 3=18$ numbers). This allready adds up to $3\frac{45(6)}{2}=405$. For the remaining $38$ we must use three numbers, we use the number $13$ two times and the number $12$ once. Therefore it is possible for $s$ to be three.
A: We have to distribute $500-443=57$ balls into $20$ containers such that the contents of the containers are as different as possible. Having two containers each with $0$, $1$, $\ldots$, $9$ balls requires $90$ balls. Therefore it is impossible to do it with no three containers getting an equal number of balls.
We now try it with three containers each getting $0$, $1$, $\ldots$ balls. It turns out that we need $3(0+1+\ldots+5)=45$ balls to serve $18$ of the $20$ containers, and that there are just $2\cdot 6$ balls left to  put into the last two containers.
