We show that the answer is $\binom{n-1}{2}+1$. To see that this is optimal, we set $a_1 = 0$ and $a_i = 2^{i-2}$ for $i \geq 2$ and observe that every pair $(i, j)$ is interesting so long as $j < n$, except for $(n-1, n)$. There are exactly $\binom{n-1}{2} + 1$ such pairs, so we have the construction.

It remains to show that in general, we cannot do better than $\binom{n-1}{2}+1$. We now show a way to find $n-2$ uninteresting pairs given any sequence $a_1 < a_2 < \dots < a_n$.

First we have the pair $(1, n)$. It is clearly uninteresting, since we certainly cannot have $a_n-a_1 \geq a_l-a_k = 2(a_n-a_1)$. Now for each $1 < i < n$ we see that unless $a_i = \frac{a_1+a_n}{2}$ one of the pairs $(1, i)$ or $(i, n)$ is uninteresting, since one of $a_i-a_1$ and $a_n-a_1$ exceeds $(a_n-a_1)/2$.

This gives us at least $1 + (n-3) = n-2$ pairs. This finishes, since now we have eliminated $n-2$ distinct pairs from a total of $\binom{n}{2}$, leaving us with exactly $\binom{n-1}{2}+1$, solving the problem.