2019/017) Each of the numbers $x_1,x_2,\cdots,x_{101}$ is $\pm1$. what is the smallest positive value of $\sum_{1 \le x_i\lt x_j\le 101 }x_ix_j$

Let the given sum be S

We have $(\sum_{1 \le  n \le 101 }x_n)^2= \sum_{1 \le  n \le 101 }x_n^2 + 2 \sum_{1 \le  x_i\lt x_j\le 101 }x_ix_j  $

so $2S = (\sum_{1 \le  n \le 101 }x_n)^2 - \sum_{1 \le  n \le 101 }x_n^2$

 as each $x_i$ is $pm1$ so $x_i^2 = 1$

so $\sum_{1 \le  n \le 101 }x_n^2 = 101$
so $2S = (\sum_{1 \le  n \le 101 }x_n)^2 - 101$
 For S to be positive we must have
$(\sum_{1 \le  n \le 101 }x_n)^2 \gt   101$ and odd
for S to be smallest we must have
$(\sum_{1 \le  n \le 101 }x_n)^2 $ smallest number greater than 101 and it is true when $(\sum_{1 \le  n \le 101 }x_n)^2 = 121$
this is possible and this give S = 10.