Mathematica competitions in Croatia

First note that the double sum of all the products $x_i{x}_j$ for $1\le i<j\le 2014$ equals $$(x_1+\cdots +x_{2014}{)}^2-(x_1^2+\cdots +x_{2014}^2).$$ Denote $A=(x_1+\cdots +x_{2014}{)}^2$ and $B=x_1^2+\cdots +x_{2014}^2$. We want to minimize $A$ and maximize $B$ at the same time. Obviously, $A\ge 0$. The minimum $A=0$ is attained when among $x_i$ there is an equal number of $1$'s and of $-1$'s, e.g. when $x_1=x_2=\cdots =x_{1007}=1$, $x_{1008}=x_{1009}=\cdots =x_{2014}=-1$. Clearly, $B=x_1^2+\cdots +x_{2014}^2\le 1^2+\cdots +1^2=2014$. The maximal value $B=2014$ is attained if none of the $x_i$ is $0$, e.g. $x_1=x_2=\cdots =x_{1007}=1$, while $x_{1008}=x_{1009}=\cdots =x_{2014}=-1$. Since the minimum of $A$ and the maximum of $B$ can be attained for the same values of $x_1,\dots ,x_{2014}$, we conclude that the minimal possible value of the sum of all products of pairs $x_i{x}_j$ ($1\le i<j\le 2014$) is $\frac{A-B}{2}=-1007$.