The accepted answer has very little to do with the question - whether or not the notion of "parity" persists $\!\bmod n.\,$ This is true $\!\iff\! n\,$ is even. Indeed by the linear congruence solvability criterion

$$\,a\,\ \text{is even }\!\bmod n\!\iff\! \exists\:\! x\!:\ 2x \equiv a\!\!\!\pmod{\! n}\! \iff\!\color{#c00}{\gcd(2,n)}\mid a\qquad$$

If $\,n\,$ is odd then $\,\color{#c00}{\gcd(2,n)=1}\,$ divides $\,a\,$ for *all* $\,a,\,$ so every integer is even (i.e. a multiple of $\:\!2).\,$ Hence there is no useful notion of parity modulo odd $\,n.$

If $\,n\,$ is even then $\,\color{#c00}{\gcd(2,n)=2}\,$ so $\,a\,$ is even $\!\bmod n\!\iff\!\color{#c00}2\mid a\!\iff\! a\,$ is even in $\,\Bbb Z$.

Further, by congruences persist mod factors of the modulus, congruences $\!\bmod 2k$ *persist* $\!\bmod 2,\,$ which means that parity arithmetic works fine $\!\bmod 2k\,$ (i.e. $\Bbb Z_{\large 2k}\,$ has $\,\Bbb Z_{\large 2}\,$ as a ring image).

More generally, as explained here, we can extend the notion of parity not only to even moduli but further to any ring having image $\,\Bbb Z_{\large 2} = \Bbb Z\bmod 2 =$ integers $\!\bmod 2,\,$ e.g. the Gaussian integers $\,\Bbb Z[i]\,$ where $\,i\,$ is odd, and $\,\Bbb Z[\sqrt 5]\,$ where $\,\sqrt 5\,$ is odd (here where we use parity in this quadratic number ring to give a simple proof that the integer $\,(9+4\sqrt{5})^n + (9-4\sqrt{5})^n\,$ is even), and here we extend parity to a ring with *infinite* elements. See also here and here for further background.