6. Proof: Let $1<a_{2}<\cdots<a_{\varphi(m)}$ be all positive integers not greater than $m$ and coprime with $m$. Since $b_{1}, b_{2}, \cdots, b_{p(m)}$ is a reduced residue system modulo $m$, and $(a, m)=1$, by Lemma 13, we know that $a b_{1}, a b_{2}, \cdots, a b_{q(m)}$ is also a reduced residue system modulo $m$. And $a b_{i} \equiv r_{i}(\bmod m), 0 \leqslant r_{i}<m$, so $r_{1}, r_{2}, \cdots, r_{q(m)}$ and $1, a_{2}, \cdots, a_{q(m)}$ may only differ in order.
Therefore,
$$r_{1}+r_{2}+\cdots+r_{q(m)}=1+a_{2}+\cdots+a_{\phi(m)}$$

By the result of Question 5,
$$1+a_{2}+\cdots+a_{\varphi(m)}=\frac{1}{2} m \cdot \varphi(m)$$

So,
$$\begin{array}{l} \frac{1}{m}\left(r_{1}+r_{2}+\cdots+r_{\varphi(m)}\right) \\ \quad=\frac{1}{m}\left(1+a_{2}+\cdots+a_{\varphi(m)}\right)=\frac{1}{2} \varphi(m) \end{array}$$