Solution:

Induction on $n$. It is obvious for $n = 3$, because the only pairs are $(1, 3)$ and $(2, 3)$, and $1 / 3 + 1 / 6 = 1 / 2$.

Now suppose it is true for $n$. As we move to $n + 1$, we introduce the new pairs $(a, n + 1)$ with $a$ relatively prime to $n + 1$ and we lose the pairs $(a, n + 1 - a)$ with $a$ relatively prime to $n + 1 - a$ and hence to $n + 1$.

So for each $a$ relatively prime to $n + 1$ and $a < (n + 1) / 2$ we gain $(a, n + 1)$ and $(n + 1 - a, n + 1)$ and lose $(a, n + 1 - a)$. But

$$\frac{1}{a(n + 1)} + \frac{1}{(n + 1 - a)(n + 1)} = \frac{n + 1 - a + a}{a(n + 1 - a)(n + 1)} = \frac{1}{a(n + 1 - a)}.$$