Note that $10$ numbers $2^0,2^1,2^2,...,2^9$ must be coloured differently. Therefore, the minimum value of $n$ is $10$, as the $i-$th color $(1 \le i \le 10)$ can be used to color numbers from $2^{i-1}$ to $2^i-1$.

i think this problem is older

this problem appeared in argentina, if I'm not mistaken.

$\color{blue} \boxed{\textbf{SOLUTION}}$ First, $1,2,4,8,..512 \equiv 2^k, 0 \leq k \leq 9$ these 10 numbers must have different colours. So, $n \geq 10$ $\color{red} \boxed{\textbf{Claim}}$ $n=10$ is possible. $\textbf{proof :}$ Consider the following numbers, $2^{k}, 2^{k}+1, ... 2^{k+1}-2, 2^{k+1}-1$ If we choose any two of these numbers, no matter what we choose, no two divides one another, as there is no divisors of anynumber in the sequence. So, we can colour all of them same! So, We if we colour all numbers of the form $2^{k}$ we can colour other numbers! Let, $j$ be a number between $1,1000,$ So, there exist some $i$ s.t. $2^{i} \leq j \leq 2^{i+1}, 1 \leq i+1 \leq 9$ We know $2^{i},2^{i+1}$ are of different colour, WLOG They are respectively Red and Blue, If, $j \in [2^{i}, 2^{i+1}-1]$ we can colour $j$ as Red from our claim above. If not $j=2^{i+1} \implies j$ is Blue! So we showed that if we colour all numbers of the form $2^k,$ we can colour any number $j$ There are total $10$ numbers of the form $2^k,$ So, $\color{red} \boxed{\textbf{minimal n=10}}$ $\blacksquare$