jmc

Let $m$ and $n$ denote the number of 1's and $-1$'s among the $a_i,$ respectively. Then $m+n=95$ and $$a_1^2+a_2^2+\cdots +a_{95}^2=95.$$Let $$S=\sum _{1\le i<j\le 95}{a}_i{a}_j.$$Then $$2S+95=(a_1+a_2+\cdots +a_{95}{)}^2=(m-n{)}^2.$$Note that $m-n=m+n-2n=95-2n$ is odd, so $(m-n{)}^2$ is an odd perfect square. To minimize $S,$ while still keeping it positive, we take $(m-n{)}^2$ as the smallest odd perfect square greater than 95, which is 121. Then $S=\frac{121-95}{2}=13.$

Equality occurs when $m=53$ and $n=42,$ so the smallest possible positive value of $S$ is $\boxed{13}.$