26th Turkish Mathematical Olympiad

Yes, Braker can guarantee that any two boxes will contain different number of balls. Suppose that Writer at the first move writes a pair $(A_1,A_2)$. At each move Braker chooses pairs containing box $A_1$ (if possible). By doing that he can guarantee that all pairs $(A_1,A_i)$, $i=2,3,\dots ,2018$ are on the table. Braker numerates all $2015$ pairs not containing $A_1$ randomly by numbers $1,2,\dots ,2015$ and accordingly puts balls into these boxes. Let $t(A_i)$ be the total number of balls in the box $A_i$ after this procedure. Without loss of generality, assume that $$t(A_2)\le t(A_3)\le \cdots \le t(A_{2018})$$ After that Braker for each $i=2,3,\dots ,2018$ numerates $(A_1,A_i)$ by $2014+i$ and accordingly distributes balls. Hereby we get $$t(A_2)<t(A_3)<\cdots <t(A_{2018}).$$ For each $i=2,3,\dots ,2018$ the box $A_i$ received balls at most $2016$ times and only in one case the number of balls was more than $2015$. Therefore, $$t(A_1)=2016+2017+\cdots +4032>2016\cdot 2016+4032>t(A_i)$$ and as a result any two boxes contain different number of balls.