NZMO Round One

Solution:
Since $a$ and $b$ are both factors of $n$, we can find positive integers $x$ and $y$ such that $a = \frac{n}{x}$ and $b = \frac{n}{y}$. Then $\frac{n}{x} + \frac{n}{y} = \frac{n}{2}$ so
$$\frac{1}{x} + \frac{1}{y} = \frac{1}{2}.$$
Without loss of generality assume $a \leqslant b$. So $x \geqslant y$ and $\frac{1}{x} \leqslant \frac{1}{y}$ and thus
$$\frac{2}{y} \geqslant \frac{1}{x} + \frac{1}{y} = \frac{1}{2}.$$
Therefore $y \leqslant 4$. So we try $y = 1, 2, 3, 4$ one by one.

Case 1 $y = 1$:
This implies $\frac{1}{x} = \frac{1}{2} - \frac{1}{y} = 0$ thus $x = -2$. This doesn't work because $a$ (and therefore $x$) must be positive.

Case 2 $y = 2$:
This implies $\frac{1}{x} = \frac{1}{2} - \frac{1}{y} = 0$ which is not possible.

Case 3 $y = 3$:
This implies $\frac{1}{x} = \frac{1}{2} - \frac{1}{y} = \frac{1}{6}$ and thus $x = 6$. This means that $n$ must be a multiple of $6$. Now let $n = 6k$, and we get
$$a = \frac{n}{x} = \frac{6k}{6} = k \qquad \mathrm{and} \qquad b = \frac{n}{y} = \frac{6k}{3} = 2k.$$
This yields the family of solutions $(a, b, n) = (k, 2k, 6k)$ where $k$ is any positive integer.

Case 4 $y = 4$:
This implies $\frac{1}{x} = \frac{1}{2} - \frac{1}{y} = \frac{1}{4}$ and thus $x = 4$. This means that $n$ must be a multiple of $4$. Now let $n = 4k$, and we get
$$a = \frac{n}{x} = \frac{4k}{4} = k \qquad \mathrm{and} \qquad b = \frac{n}{y} = \frac{4k}{4} = k.$$
This yields the family of solutions $(a, b, n) = (k, k, 4k)$ where $k$ is any positive integer.

Remember that we assumed $a \leqslant b$ so we still need to swap the roles of $a$ and $b$ for our final answer. So in summary, the triples $(a, b, n)$ which satisfy the given equation are
$$(k, 2k, 6k),\ (2k, k, 6k)\ \text{and}\ (k, k, 4k)$$
where $k$ is any positive integer.