Let $c_1,c_2,\ldots,c_{\varphi(m)}$ be the reduced residue set modulo $m>2$. Show that $$c_1+c_2+\cdots+c_{\varphi(m)} \equiv 0 \pmod{m}.$$
My solution looks something like this.
If $c_i \in {\mathbb{Z}_m}^*$, then $m-c_i \in {\mathbb{Z}_m}^*$. Hence, we may take the reduced residue system $$\{c_1,c_2,\ldots,c_{\varphi(m)/2},m-c_{\varphi(m)/2},\ldots,m-c_2,m-c_1\}.$$
Hence, \begin{align} c_1+c_2+\cdots+c_{\varphi(m)} &= c_1+c_2+\cdots+c_{\varphi(m)/2}+m-c_{\varphi(m)/2}+\cdots+m-c_2+m-c_1 \\ &=m \cdot \varphi(m)/2 \equiv 0 \pmod{m}, \end{align}
since $\varphi(m)$ is even for $m >2$.
Is this on the right track?
