6. Given 95 numbers $a_{1}, a_{2}, \cdots, a_{95}$, each of which can only take one of the two values +1 or -1, then the sum of their pairwise products
$$
a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{94} a_{95}
$$

is the smallest positive value . $\qquad$
6. Note
$$
N=a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{94} a_{95}
$$

Let $a_{1}, a_{2}, \cdots, a_{95}$ have $m$ of $+1$ and $n$ 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
$$
\begin{array}{ll} 
& \left(a_{1}+a_{2}+\cdots+a_{95}\right)^{2}=2 N+95 . \\
\text { Also } \quad & a_{1}+a_{2}+\cdots+a_{95}=m-n, \\
\therefore \quad & (m-n)^{2}=2 N+95 .
\end{array}
$$

The smallest natural number $N=13$ that satisfies the above equation, at this point $(m-n)^{2}=11^{2}$, i.e.,
$$
m-n= \pm 11 \text {, }
$$

Solving (2) and (3) simultaneously gives $m=53, n=42$ or $m=42, n=53$, from which we can construct the case where $N$ reaches its minimum value.
Therefore, the smallest positive value is 13.