6. Given 95 numbers $a_{1}, a_{2}, a_{3}, \cdots, a_{95}$, each of which can only take one of the two values +1 or -1, then the minimum value of the sum of their pairwise products $a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{94} a_{95}$ is $\qquad$ .
13
6. 【Analysis and Solution】Let $N=a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{94} a_{95}$.

Suppose $a_{1}, a_{2}, \cdots, a_{95}$ have $m$ numbers of $+1$ and $n$ numbers of -1, then $m+n=95$.
Multiplying (1) by 2 and adding $a_{1}^{2}+a_{2}^{2}+\cdots+a_{95}^{2}=95$, we get
$$
\begin{array}{l}
\left(a_{1}+a_{2}+\cdots+a_{95}\right)^{2}=2 N+95 . \\
\text { Also } a_{1}+a_{2}+\cdots+a_{95}=m-n, \\
\therefore(m-n)^{2}=2 N+95 .
\end{array}
$$

The smallest natural number $N=13$ that satisfies the above equation, at this time $(m-n)^{2}=11^{2}$,
i.e., $m-n= \pm 11$,
Solving (2) and (3) simultaneously gives $m=53, n=42$ or $m=42, n=53$, according to which we can construct the case where $N$ reaches its minimum value. Therefore, the smallest positive value sought is 13.