Solution
= Solution
Let $R_n=\{k\in\mathbb Z:0<k<n,\ \gcd(k,n)=1\}$. The map $k\mapsto n-k$ sends $R_n$ to itself because
$$
\gcd(n,n-k)=\gcd(n,k)=1.
$$
Applying it twice returns $k$, so it is a bijection. Reindexing the finite sum by this bijection gives
$$
\boxed{\sum_{k\in R_n}k=\sum_{k\in R_n}(n-k).}
$$
There is no need to assume that every residue belongs to a pair of distinct elements; the reindexing also handles the fixed point when $n=2$.