Harvard-MIT Mathematics Tournament

Solution:
Suppose $m$ is the minimum integer for which the given configuration $C$ on set $A$ is $m$-separable, and fix a corresponding labeling of the elements of $A$. Let $A_{i}$ be the set of all elements with the label $i$. Then, for any $i, j$ with $1 \leq i < j \leq m$, there must exist $a_{i} \in A_{i}, a_{j} \in A_{j}$ with $\{a_{i}, a_{j}\} \in C$, since otherwise the elements of $A_{j}$ could have been reassigned the label $i$, decreasing the number of distinct labels necessary and thus contradicting the minimality of $m$.

We thus get at least $\binom{m}{2}$ different elements of $C$. Therefore, $e \geq \binom{m}{2} = \frac{m(m-1)}{2} = \frac{\left(m-\frac{1}{2}\right)^2 - \frac{1}{4}}{2}$, and solving for $m$ gives the desired result.