A Problem That Requires No More Than Some Simple Logic.

image containing math problem statement
This problem can be tackled using some simple logical analysis. No special knowledge of mathematics is required.
Let:
$S = \displaystyle \sum\limits_{1 \le i \le j \le 2024} a_ia_j$

$\Rightarrow 2S =  \displaystyle 2 \cdot \left(\sum\limits_{1 \le i \le j \le 2024} a_ia_j \right)$

-----------book page break-----------
$\Rightarrow 2S = \displaystyle \left(\sum\limits_{1 \le i \le j \le 2024} 2 \cdot a_ia_j \right)$

$\Rightarrow 2S = (a_1 + a_2 + a_3 + ... + a_{2024})^2 - (a_1^2 + a_2^2 + a_3^2 + ... + a_{2024}^2)$

For any $a_i = \pm 1$, $a_i^2 = 1$,
$\therefore (a_1^2 + a_2^2 + a_3^2 + ... + a_{2024}^2) = 2024$

Therefore,

$2S = (a_1 + a_2 + a_3 + ... + a_{2024})^2 - 2024$            $...(i)$
Therefore, the smallest positive value of $2S$ is achieved when $(a_1 + a_2 + a_3 + ... + a_{2024})^2$ is the smallest even perfect square greater than $2024$.

Let the count of $(+1)$s in the series be $m$, therefore the count of $(-1)$s is $2024 - m$.

Then the sum of all the terms will be:
$a_1 + a_2 + a_3 + ... + a_{2024} = m \times (+1) + (2024 - m)(-1)$

$\Rightarrow a_1 + a_2 + a_3 + ... + a_{2024} = 2m - 2024$

$\Rightarrow (a_1 + a_2 + a_3 + ... + a_{2024})^2 = (2m - 2024)^2$

Substituting this value in $eqn\ (i)$, we get:
$2S = (2m - 2024)^2 - 2024$

Observe that $2m - 2024$ will always be an even number.


-----------book page break-----------
The smallest even perfect square, greater than $2024$ is $46^2 = 2116$

Therefore, the smallest positive value of $2S$ is:
$2S = 2116 - 2024 = 92$

$\Rightarrow S = 46$