Relevant

Can one do it with elementary methods, i.e, without considering Farey sequences?

Let $A_n$ be the desired sum. Remark that $A_1 = \tfrac{1}{2}$ and \begin{align*} A_{n} - A_{n-1} &= \sum_{\substack{\gcd(a,n) = 1, \\ a < n}} \frac{1}{na} - \sum_{\substack{a+b=n, a<b \\ \gcd(a,b) = 1}} \frac{1}{ab} \\ &= \frac{1}{n} \sum_{\substack{\gcd(a,n) = 1, \\ a < n}} \frac{1}{a} - \sum_{\substack{\gcd(a,n) = 1 \\ a \le n/2}} \frac{1}{a(n-a)} \\ &= \frac{1}{n} \sum_{\substack{\gcd(a,n) = 1, \\ a < n}} \frac{1}{a} - \frac{1}{n} \sum_{\substack{\gcd(a,n) = 1 \\ a \le n/2}} \left( \frac{1}{a} + \frac{1}{n-a} \right) \\ &= \frac{1}{n} \sum_{\substack{\gcd(a,n) = 1, \\ a < n}} \frac{1}{a} - \frac{1}{n} \sum_{\substack{\gcd(a,n) = 1 \\ a \le n}} \frac{1}{a} \\ &= 0. \end{align*}