Test D 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 the sum of their pairwise products $a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{94} a_{95}$ is $\qquad$ .
(1994 National High School League)
Let $N=a_{1} a_{2}+a_{1} a_{3}+\cdots+a_{94} a_{95}$. Suppose $a_{1}, a_{2}, \cdots, a_{95}$ contain $m$ +1s and $n$ -1s. Then $m+n=95$, hence we have
$$
\left(a_{1}+a_{2}+\cdots+a_{95}\right)^{2}=2 N+95,
$$

and
$$
\left(a_{1}+a_{2}+\cdots+a_{95}\right)=m-n,
$$

thus
$$
(m-n)^{2}=2 N+95.
$$

The smallest natural number $N=13$ that satisfies this equation gives $(m-n)^{2}=11^{2}$. Therefore, $m-n = \pm 11$. Solving the system $m+n=95, m-n= \pm 11$, we get $m=53, n=42$, or $m=42, n=53$. This allows us to construct the values of $a_{1}, a_{2}, \cdots, a_{95}$ that achieve the minimal value of $N=13$.