65th Czech and Slovak Mathematical Olympiad

Let $s$ be the sum, $n$ the number and $p$ the integer part of the mean of the numbers on the board. Then we can write $$\frac{s}{n}=p+\frac{2,016}{10,000}=p+\frac{126}{625},$$ which gives $$625(s-pn)=126n.$$ Numbers $126$ and $625$ are coprime, thus $625\mid n$. Therefore $n\ge 625$. The numbers on the board are different, that is $$p=\frac{s}{n}-\frac{126}{625}\ge \frac{1+2+\cdots +n}{n}-\frac{126}{625}=\frac{n(n+1)}{2n}-\frac{126}{625}=\frac{n+1}{2}-\frac{126}{625}\ge \frac{625+1}{2}-\frac{126}{625}>312.$$ The integer $p$ is thus at least $313$ and the value of the mean at least $313,2016$. This value can be attained by numbers $1,2,\dots ,624$ and $751$. We get $$\frac{1+2+\cdots +624+751}{625}=\frac{312\cdot 625+751}{625}=313+\frac{126}{625}=313,2016.$$