If $z \in \Bbb{Z}/n$, w let $z^* =$ the standard residue or in other words the least non-negative integer equal to $z$ modulo $n$. Suppose that $n \mid m$ for some two positive integers $n,m$.
If $\pi : \Bbb{Z}/m \twoheadrightarrow \Bbb{Z}/n$ is the ring surjection, it's always true that:
$$ \left\lfloor\frac{x - \pi(z)^*}{n} \right\rfloor \geq \left\lfloor\frac{x - z^*}{m}\right\rfloor $$
for all $x \in \Bbb{Z}$ and $z \in \Bbb{Z}/m$. In particular, if we must use this, my application is such that $z$ solves $z^2 = 1\mod m$.
If $n \lt m$ in addition, can we say that the conjectured inequality is strict as well?
Also, in my application $n,m$ are always square-free. Use that if it needs it.
Edit. Please also assume that $x \geq 0$. Because I just now found a proof, and I'll post as an answer, and $x\geq 0$ is suitable to my app as well.
It's true for $x = 0$ because both floors evaluate to $-1$, as $r^* \lt m$ and $\pi(r)^* \lt n$ (as they are the standard residues).
Now assume that it's true for all $x = 0...X$. Then at $x = X + 1$ we have:
$$ \lfloor \frac{ X + 1 - r^*}{m}\rfloor = \lfloor\frac{ X - r^*}{m}\rfloor + (m \mid X + 1 -r^*) $$
where $(m \mid z)$ simply takes on the value $1$ if $m$ divides $z$ and $0$ otherwise.
Similarly goes for the other side of the original inequality. We have by inductive assumption that:
$$ \lfloor \frac{X - r^*}{m}\rfloor \leq \lfloor\frac{X-\pi(r)^*}{n} \rfloor $$
Now simply add to that:
$$ (m \mid X + 1 - r^*) \leq (n\mid X + 1 - \pi(r)^*) $$
which we will now prove. If the LHS is $1$ then:
$$ X + 1 - r^* = 0 \pmod m \\ \implies X + 1 - \pi(r)^* = X + 1 - r^* = 0 \pmod n $$
And so the RHS is also one. This completes the proof.
$\blacksquare$