Let $A=\left \{ 1,2,3,...,1000 \right \}$ be a set. Determine the greatest number $m$, so that for any $m$ numbers removed from $A$, there will exist $a,b\in A$, with $a|b$.
The only useful thought I had was that if we remove the numbers from $1$ to $500$, we won't find two numbers from $501$ to $1000$, with one of them dividing another. Hence, $m\leq 499$.
$m = 499$. You establish in your question that $m$ cannot exceed $499$. So, for $499$ to be the greatest possible value of $m$, we just need to show that it works. Sure enough, taking out $499$ elements leaves $501$ behind, and "If more than half of the integers from {${1, 2, ..., 2n}$} are selected, then some two of the selected integers have the property that one divides the other."
I'll drop the proof here in case the link ever dies:
[Sharygin]. In the set {${1, 2, ..., 2n}$} there are $n$ even and $n$ odd integers. Let $A$ consist of at least $n + 1$ numbers from that set: $|A| > n$. Every integer is a product of a power of $2$ and an odd integer. Remove the powers of two from the members of $A$. The resulting set $B$ consists of odd integers and, in addition, $|B| = |A| > n$. The terms of $B$ are among the $n$ odd members of the set {${1, 2, ..., 2n}$}, meaning that some two of them must be equal. Of the corresponding terms of A, the smaller divides the larger, for the two are in the form $(2^k)b$, $(2^m)b$, with $b$ odd and $k ≠ m$.
Reference: I. F. Sharygin, Mathematical Mosaic, Mir, 2002, problem 65.3 (in Russian)