Let $S$ be a set with 2002 elements, and let $N$ be an integer with $0 \le N \le 2^{2002}$. Prove that it is possible to color every subset of $S$ either blue or red so that the following conditions hold:
(a) the union of any two red subsets is red;
(b) the union of any two blue subsets is blue;
(c) there are exactly $N$ red subsets.
What does the phrase coloring of subsets mean? I also found a link which provides the solution
https://artofproblemsolving.com/wiki/index.php?title=2002_USAMO_Problems/Problem_1
In this solution why do we consider the case where $N \geq 2^k$. N equal to $2^k$ is fine, but isn't N an integer less than $2^k$ since in the solution we proceed by induction to generalise it for any integer n, not only 2002?
In the inductive step we assume we have proven that for any set of size $k$ we can properly color the subsets so that $N$ of them are red for any $N$ in the range $[0,2^k]$. We now want to prove the same for $k+1$, which means there are $2^{k+1}$ subsets and we need to show that we can properly color them for any $N$ in the range $[0,2^{k+1}]$. We divide the set into $\{s\}$, the new element and $S'$, which is everything else and has $k$ elements.
The proof then says essentially, if $N \le 2^k$ we had a proper coloring of $S'$ with $N$ subsets colored red, so color all the subsets that include $s$ blue and we are done. If $N \gt 2^k$ you can take all the subsets that include $s$, which is $2^k$ of them, color them red, and fill out with a proper coloring of the set with $k$ elements that has $N-2^k$ red subsets.