NZMO Round One

Solution:
Each set $A_{i}$ has exactly three pairs of elements. But each unordered pair chosen from $\{1,2,\ldots, n\}$ can be in at most one such set. Therefore
$$\binom{n}{2} \geq 3m.$$
This establishes the required upper bound on the size of $m$.

Let $T$ be the set of all triples, i.e.
$$T = \{(a, b, c) \mid 1 \leq a < b < c \leq n\}.$$
We now partition $T$ into $n$ parts, $T = T_{0} \cup T_{1} \cup T_{2} \cup \dots \cup T_{n-1}$ based on the residue of $a + b + c$ modulo $n$, i.e.
$$T_{i} = \{(a, b, c) \mid (a, b, c) \in T \mathrm{~and~} a + b + c \equiv i \mathrm{~(mod~} n)\}$$
So we have $\binom{n}{3}$ triples (our pigeons) and $n$ pigeonholes (the parts $T_{0}, T_{1}, T_{2}, \ldots, T_{n-1}$) therefore by the pigeonhole principle at least one part $T_{k}$ must have at least
$$\frac{1}{n} \times \binom{n}{3}$$
triples in it. Now it suffices to show that $T_{k}$ satisfies the problem. For the sake of contradiction, assume there exists two triples $(a, b, c), (a', b', c') \in T_{k}$ such that $a = a'$ and $b = b'$ but $c \neq c'$. This would imply
$$a + b + c \equiv j \equiv a' + b' + c' \pmod{n}.$$
Hence $c \equiv c'$ (mod $n$) and thus $c = c'$ and so we get $(a, b, c) = (a', b', c')$, contradiction. Therefore no such pair $(a, b, c), (a', b', c') \in T_{k}$ exists. Therefore, letting the elements of $T_{k}$ be our sets $A_{i}$, we achieve
$$m \geq \frac{1}{n} \times \binom{n}{3} = \frac{(n - 1)(n - 2)}{6}$$
as required.