Solution:

We use induction on $n$. If $n = 2$, then the only such fraction is $r = 1$, $s = 2$, giving $1/rs = 1/2$, so the result holds.

Suppose it holds for $n - 1$. As we move to $n$, we lose the fractions with $r + s = n$. The other fractions $1/rs$ which satisfy the conditions for $n - 1$ also satisfy the conditions for $n$. We also gain the fractions with $s = n$. These have sum $= 1/n$ (sum $1/r$ for all $r$ satisfying $0 < r < n$ and $r$ relatively prime to $n$).

But if $r$ is relatively prime to $n$, then so is $n - r$, and $n - r$ does not equal $r$ (otherwise $r$ divides $n$). The pair $1/r$, $1/(n - r)$ has sum $n/(r(n - r))$. So the fractions with $s = n$ have sum equal to the sum of all $1/(r(n - r))$ with $0 < r < n$ and $r$ relatively prime to $n$. But that is exactly the sum of the fractions lost. Thus the total is unchanged as we move from $n - 1$ to $n$.