Show that if $m$ is odd then $\frac{(-m+1)}{2}, \frac{(-m+3)}{2}, \cdots, \frac{(m-3)}{2}, \frac{(m-1)}{2}$ is a complete residue system modulo m.
I understand the definition of a complete residue system but is having a hard time incorporating the odd part of this statement.
If $r_0, \ldots, r_{m-1}$ is a complete residue system modulo $m$ and $k$ is an integer, then $r_0-k, \ldots, r_{m-1}-k$ is a complete residue system modulo $m$ as well.
Note that $k=\frac{m-1}{2}$ is an integer as $m$ is an odd number. Since $0, 1, \ldots, m-1$ is obviously a complete residue system modulo $m$ we conclude that $$ \frac{(-m+1)}{2}=0-\frac{m-1}{2},\frac{(-m+3)}{2}=1-\frac{m-1}{2}, \ldots, \frac{(m-1)}{2}=m-1-\frac{m-1}{2}$$ is a complete residue system modulo $m$, as well.
Let $\,m = 2n\!+\!1\,$ so $\,(m\!-\!1)/2 = n,\,$ Your sequence is $\,-n,-(n\!-\!1),\ldots,-1,0,1,\ldots,n\!-\!1,n,\,$ which is a sequence of $\,m = 2n+1\,$ consecutive integers. It is just the standard complete residue system shifted left by $\,n,\,$ i.e. mapped by $\,x\mapsto x-n,\,$ which is a bijection mod $\,m.\,$
Remark $ $ This system is sometimes preferred since it may introduce simplifications, e.g. the smaller numbers may simplify calculations as here, and it may help exploit innate reflection (negation) symmetry, e.g. in this additive form of Wilson's theorem (sum of all elements $\!\bmod n$).