6. (1994 National High School Competition Question) Given 95 numbers $a_{1}, a_{2}, \cdots, a_{95}$, each number can only take +1 or -1, then the minimum value of the sum of their pairwise products $\sum_{1 \leq i<j \leq 95} a_{i} a_{j}$ is
6. 13. Let $N=\sum_{1<i<j \leqslant 95} a_{i} a_{j}, a_{1}, a_{2}, \cdots, a_{95}$ have $m$ values of $+1$ and $n$ values of -1, then $a_{1}^{2}+a_{2}^{2}+\cdots+a_{95}^{2}=95, m+n=95, a_{1}+a_{2}+\cdots+a_{95}=m-n$. Therefore, $(m-n)^{2}=\sum_{i=1}^{95} a_{i}^{2}+2 \sum_{1 \leqslant i<j \leqslant 95} a_{i} a_{j}=95+2 N$. The smallest positive integer $N=13$ that makes $95+2 N$ a perfect square. At this time, $(m-n)^{2}=11^{2}, m-n= \pm 11$ and $m+n=95$, solving these equations gives $m=53, n=42$ or $m=42, n=53$. Hence, the smallest positive value is 13.