PAMO 2015 question 1 Let $a_1,a_2,...,a_{11}$ be integers. 
Prove that there are numbers $b_1,b_2,...,b_{11}$,
each $b_i$ equal $-1,0$ or $1$, but not all being $0$, such that the number
$N = a_1b_1 +a_2b_2 +···+a_{11}b_{11}$
is divisible by $2015$.
The question sems so ambiguous, I don't even know where to start. I'd appreciate any hints or a full solution.

Consider all the sums $\sum_{i=1}^{11}a_ix_i$, where $x_i \in \{0,1\}$. There are $2^{11} > 2015$ of these sums, so two of them have the same remainder modulo $2015$. Then we take the difference, which has the required form.
