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Let's say I'm given $\binom{n}{2}$ pairs of numbers. What is the maximum number of triples (there are $\binom{n}{3}$ of them) I can form by using 3 pairs for every triplet? For example, I can use $(2,5),(5,8),(8,2)$ to form the triplet $(2,5,8)$. Remember that I can only use each pair once.

I've tried systematically counting, first counting each triplet with 1 in it, then 2, etc. but I found it to be very messy.

What's a good way to do this?

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To use them all you are asking for a Steiner triple system $(2,3,n)$ These exist whenever $n\equiv 1,3 \pmod 6$

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    It could be for small numbers because they interfere. – 2017-01-17
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    What would be the maximum if $n$ was not $1,3\mod6$? – 2017-02-22