Croatian Mathematical Competitions

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 $\frac{A-B}{2} = -1007$.