Let $n$ be a positive integer, and consider the set $\{1, \dots, n\}$. If we remove from this set all the numbers $a$ which satisfy
$$
a \equiv 0 \mod d
$$
for at least one divisor $d$ of $n$ (where we only consider $d \neq 1$, of course), then it is well-known that the total number of remaining elements is given by the Euler totient function $\varphi(n)$. This is quite easy to show - note that it uses the fact that actually it is sufficient to consider those divisors $d$ which are *primes*, since then all composite divisors are incorporated automatically.

Now here is my problem. Let $\gamma \in [0,1]$ be a fixed real number. For a real number $y$, let $[y]$ denote the integer which is closest to $y$. From the set $\{1, \dots, n\}$ I have to remove all the elements $a$ for which $$ a \equiv [\gamma d] \mod d $$ for at least one divisor $d$ of $n$ (where again we only consider $d \neq 1$). How many elements remain? In contrast to the case $\gamma=0$, now it obviously is not sufficient anymore to consider only prime divisors $d$. A trivial lower bound for the number of remaining elements is $$ n - \sum_{d | n, ~d \neq 1} \frac{n}{d} = n - \sum_{d | n, ~d \neq n} d, $$ since for every divisor $d$ we remove at most one residue class (which contains $n/d$ elements).

Question: Is there any significantly better lower bound for the number of remaining elements? (Note that I do not want to assume anything whatsoever on $n$ or $\gamma$.)