Team selection tests for BMO 2018

First, we show that the sum is at most $1008$ and at least $-1008$. Indeed, suppose that there are $a$ numbers $+1$ in $2018$ given numbers and $b$ products equal $-1$. Then the sum of $2018$ given numbers is $a+(-1)(2018-a)=2a-2018$, and sum of $2018$ products is $$(2018-b)+(-1)b=2018-2b.$$ Since $2018-2b<0$ we have $b\ge 1010$. And because each product of $-1$ comes from one $1$ and one $-1$, and one $-1$ can contribute to at most two $-1$ products. Hence the number of $-1$ numbers is at least $1010/2=505$, or $2018-a\ge 505$. Hence $a\le 1513$, or $2a-2018\le 1008$. Similarly, each product of $-1$ comes from one $1$ and one $-1$, and one $1$ can contribute to at most two $-1$ products. Hence the number of $1$ numbers is also at least $505$. Hence, $a\ge 505$, or $2a-2018\ge -1008$. Combine these results, we showed that the sum is in range $[-1008,1008]$. It is clear that the sum is even, and we will show that the sum can be any even number in that range. We start from a configuration of $1009$ numbers of $+1$ and $1009$ numbers of $-1$ alternating around the circle. In this case, the sum of $2018$ numbers is $0$ and the sum of the product is $-2018$. On one way, we eventually change one $-1$ to one $1$ exactly $504$ times. Each time, the sum of $2018$ numbers increases by $2$, so the sum will increase from $0$ to $1008$. Besides, the sum of $2018$ products will increase by $4$ (two $-1$ products change to two $1$ products). Hence the sum of $2018$ products is at most $-2018+504\times 4<0$. On another way, we eventually change one $+1$ to one $-1$ exactly $504$ times. By doing these steps, the sum of $2018$ numbers will decrease from $0$ to $-1008$ and the sum of $2018$ products is always negative. Therefore, all even numbers from $-1008$ to $1008$ are obtainable.