12. 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 minimum positive value of their pairwise products' sum $a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{94} a_{95}$ is $\qquad$ .
$$
\begin{array}{l}
2\left(a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{94} a_{95}\right)=\left(a_{1}+a_{2}+\cdots+a_{95}\right)^{2}-\left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{95}^{2}\right) \\
=\left(a_{1}+a_{2}+\cdots+a_{95}\right)^{2}-95 \geqslant 11^{2}-95=26 \Rightarrow a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{94} a_{95} \geqslant 13
\end{array}
$$

When the equality holds, $a_{1}, a_{2}, \cdots, a_{95}$ have 53 ones and 42 negative ones, or 53 negative ones and 42 ones. Therefore, the minimum positive value of $a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{94} a_{95}$ is 13.