SMT 2018 Problem

First, realize that:

$(\sum\limits_{i=1}^{2018} a_i)^2 = \sum\limits_{i=1}^{2018} a_i^2 + 2(\sum\limits_{1 \le i < j \le 2018} a_ia_j)$

Since $a_i = \pm 1$ , $a_i^2 = 1$ , so $\sum\limits_{i=1}^{2018} a_i^2 = 2018$ .

Let $r$ and $s$ be the number of values of $k$ where $a_k = 1$ and $a_k = -1$ , respectively. $\sum\limits_{i=1}^{2018} a_i = r(1) + s(-1) = r - s$ . Since $r + s =2018$ , $r$ and $s$ have the same parity, so $r - s$ should be an even integer $2n$ . Plugging this all into the first equation, we get:

$(2n)^2 = 2018 + 2(\sum\limits_{1 \le i < j \le 2018} a_ia_j) \implies \sum\limits_{1 \le i < j \le 2018} a_ia_j = \frac{(2n)^2 - 2018}{2}$

Since this value must be positive, and the goal is to minimize the value, $(2n)^2$ should be the smallest even perfect square greater than $2018$ , which is $46^2 = 2116$ . The answer is $\frac{2116-2018}{2} = \boxed{49}$ .

Notice that:

$2\sum\limits_{1 \le i < j \le 2018} a_ia_j = (a_1+a_2+\dots+a_{2018})^2-(a_1^2+a_2^2+\dots+a_{2018}^2) = (a_1+a_2+\dots+a_{2018})^2-2018$ .

We know that $a_k=\pm 1\text { and } a_1+a_2+\dots+a_{2018}$ is an integer between $-2018$ and $2018$ incliusive. Also, there are even number of terms where each term is odd, thus their sum is even.

The minimum positive integer value of $(a_1+a_2+\dots+a_{2018})^2 - 2018 = 46^2 -2018 =98$ . This is true when $a_1+a_2+\dots+a_{2018}=\pm 46$ which can be only accomplished when $a_1=a_2=\dots=a_{46}=1.$

How about the rest of the terms ( $a_{47}, a_{48}, a_{49},\dots, a_{2018}$ )? They can contain values between $+1$ 's and $-1$ 's.

Therefore, the least positive value is $\frac{98}{2}=\boxed{49}$