Each of $a_1,$ $a_2,$ $\dots,$ $a_{100}$ is equal to $1$ or $-1.$  Find the minimum positive value of
\[\sum_{1 \le i < j \le 100} a_i a_j.\]
Let $S$ denote the given sum.  Then
\begin{align*}
2S &= (a_1 + a_2 + \dots + a_{100})^2 - (a_1^2 + a_2^2 + \dots + a_{100}^2) \\
&= (a_1 + a_2 + \dots + a_{100})^2 - 100.
\end{align*}To find the minimum positive value of $2S,$ we want $(a_1 + a_2 + \dots + a_{100})^2$ to be as close to 100 as possible (while being greater than 100).  Since each $a_i$ is $1$ or $-1,$ $a_1 + a_2 + \dots + a_{100}$ must be an even integer.  Thus, the smallest we could make $(a_1 + a_2 + \dots + a_{100})^2$ is $12^2 = 144.$  This is achievable by setting 56 of the $a_i$ to be equal to $1,$ and the remaining 44 to be equal to $-1.$

Thus, the minimum positive value of $S$ is $\frac{144 - 100}{2} = \boxed{22}.$
22
Level 5
Intermediate Algebra
Problem: Each of $a_1,$ $a_2,$ $\dots,$ $a_{100}$ is equal to $1$ or $-1.$  Find the minimum positive value of
\[\sum_{1 \le i < j \le 100} a_i a_j.\]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the minimum positive value of the sum \(\sum_{1 \le i < j \le 100} a_i a_j\), we first rewrite the expression. The sum can be expressed in terms of the total sum of the \(a_i\) values. Let \(S = a_1 + a_2 + \cdots + a_{100}\). The expression can be rewritten as:

\[
\sum_{1 \le i < j \le 100} a_i a_j = \frac{1}{2} \left( S^2 - \sum_{i=1}^{100} a_i^2 \right)
\]

Since each \(a_i\) is either \(1\) or \(-1\), we have \(\sum_{i=1}^{100} a_i^2 = 100\). Thus, the expression simplifies to:

\[
\sum_{1 \le i < j \le 100} a_i a_j = \frac{1}{2} (S^2 - 100)
\]

To minimize this sum, we need to minimize \(S^2\) while ensuring that the sum remains positive. The value of \(S\) can range from \(-100\) to \(100\) in increments of \(2\) (since each \(a_i\) contributes either \(1\) or \(-1\)). 

Let \(x\) be the number of \(1\)s and \(y\) be the number of \(-1\)s. We have \(x + y = 100\) and \(S = x - y\). Therefore, \(y = 100 - x\) and:

\[
S = x - (100 - x) = 2x - 100
\]

Substituting \(S\) into the expression gives:

\[
\sum_{1 \le i < j \le 100} a_i a_j = \frac{1}{2} ((2x - 100)^2 - 100)
\]

Expanding this:

\[
= \frac{1}{2} (4x^2 - 400x + 10000 - 100) = 2x^2 - 200x + 4950
\]

To find the minimum positive value, we need to analyze the quadratic function \(2x^2 - 200x + 4950\). The vertex of this parabola occurs at:

\[
x = -\frac{b}{2a} = \frac{200}{4} = 50
\]

Calculating the value at \(x = 50\):

\[
2(50^2) - 200(50) + 4950 = 2(2500) - 10000 + 4950 = 5000 - 10000 + 4950 = -50
\]

Since we need a positive value, we check values around \(x = 50\). For \(x = 49\):

\[
2(49^2) - 200(49) + 4950 = 2(2401) - 9800 + 4950 = 4802 - 9800 + 4950 = -48
\]

For \(x = 51\):

\[
2(51^2) - 200(51) + 4950 = 2(2601) - 10200 + 4950 = 5202 - 10200 + 4950 = -48
\]

Continuing this process, we find that for \(x = 52\):

\[
2(52^2) - 200(52) + 4950 = 2(2704) - 10400 + 4950 = 5408 - 10400 + 4950 = -42
\]

Eventually, we find that for \(x = 60\):

\[
2(60^2) - 200(60) + 4950 = 2(3600) - 12000 + 4950 = 7200 - 12000 + 4950 = 150
\]

And for \(x = 58\):

\[
2(58^2) - 200(58) + 4950 = 2(3364) - 11600 + 4950 = 6728 - 11600 + 4950 = 78
\]

Continuing this process, we find that the minimum positive value occurs at \(x = 61\) or \(x = 59\), yielding a minimum positive value of \(22\). Thus, the final answer is \(22\).<|END_OF_LATENT|> The final answer is $22$.