Saudi Arabian Mathematical Competitions, 2012

Since $n$ is odd,
$$|S| = \phi(n) = n \prod_{p|n} (1 - 1/p)$$
is even. Given an element $k$ of $S$, write
$k^{|S|} \equiv 1+nr_k \pmod{n^2}$ and $(n-k)^{|S|} \equiv 1+nr_{n-k} \pmod{n^2}$, and notice that
$$(n-k)^{|S|} \equiv k^{|S|} \cdot n \cdot k^{|S|-1} \pmod{n^2},$$
to get
$$r_k - r_{n-k} \equiv |S| \cdot k^{|S|-1} \pmod{n}.$$
Hence
$$\prod_{k \in T} (r_k - r_{n-k}) \equiv |S|^{|T|} \left(\prod_{k \in T} k\right)^{|S|-1} \pmod{n}.$$
Finally, notice that the product in the right-hand member is congruent to 1 modulo $n$. To see this, let $k'$ denote the modulo $n$ multiplicative inverse of an element $k$ of $S$, and notice that
$$k'+1 \equiv k'(k+1) \pmod{n} \quad \text{and} \quad (k-1)(k'+1) \equiv k-k' \pmod{n}.$$
The first congruence shows that if $k$ belongs to $T$, then so does $k'$, and the second shows that $k = k'$ if and only if $k = 1$. Consequently, if $k \neq 1$, the factors $k$ and $k'$ in the product can be paired off and the conclusion follows.