PRMO 2019

Suppose that $M \ge 5$ . If we have $N = 2M+1$ numbers $x_!,xx_2,x_3,...,x_N$ , all equal to $2$ or $-2$ , then we note that $\left(\sum_{j=1}^N x_j\right)^2 \; = \; \sum_{j=1}^N x_j^2 + 2\sum_{1 \le i < j \le N}x_ix_j \; = \; 4N + 2\sum_{1 \;e i < j \le N}x_ix_j$ so that $S \;= \; \sum_{1 \le i < j \le N}x_ix_j \; = \; \frac12\left[\left(\sum_{j=1}^N x_j\right)^2 - 4N\right]$ If the numbers $x_1,x_2,x_3,...,x_N$ consist of $M+k+1$ copies of $2$ and $M-k$ copies of $-2$ , then $\sum_{j=1}^N x_j \,=\, 2(2k+1)$ , and hence we deduce that $S \; = \; \sum_{1 \le i < j \le N}x_ix_j \; = \; 2\big[(2k+1)^2 - N\big]$ where $-M-1 \le k \le M$ . Thus the smallest positive value of $S$ is equal to $2(K^2 - N)$ , where $K$ is the smallest odd positive integer such that $K^2 > N$ . Note that $M^2 > 2M+1$ and $(M-1)^2 > 2M+1$ whenever $M \ge 5$ , so that the smallest odd positive integer $K$ with $K^2 > N$ lies in the acceptable range $-M-1 \le K \le M$ .

In this case, with $N = 2019$ , the smallest positive value of $S$ is $2(45^2 - 2019) = \boxed{12}$ .