Problem: Let $S$ be a set with $2002$ elements, and let $N$ be an integer with $0 \leq N \leq 2^{2002}$.
Prove that it is possible to color every subset of $S$
either black or white so that the following conditions hold:
\begin{enumerate}
  \ii[(a)] the union of any two white subsets is white;
  \ii[(b)] the union of any two black subsets is black;
  \ii[(c)] there are exactly $N$ white subsets.
\end{enumerate}

Solution: We will solve the problem with $2002$
replaced by an arbitrary integer $n \ge 0$.
In other words, we prove:
\begin{claim*}
  For any nonnegative integers $n$ and $N$ with $0 \le N \le 2^n$,
  it is possible to color the $2^n$ subsets
  of $\{1, \dots, n\}$ black and white
  satisfying the conditions of the problem.
\end{claim*}
The proof is by induction on $n$.
When $n = 1$ the problem is easy to do by hand,
so this gives us a base case.

For the inductive step, we divide into two cases:
\begin{itemize}
  \ii If $N \le 2^{n-1}$,
  then we take a coloring of subsets of $\{1, \dots, n-1\}$
  with $N$ white sets;
  then we color the remaining $2^{n-1}$ sets
  (which contain $n$) black.

  \ii If $N > 2^{n-1}$,
  then we take a coloring of subsets of $\{1, \dots, n-1\}$
  with $N - 2^{n-1}$ white sets;
  then we color the remaining $2^{n-1}$ sets
  (which contain $n$) white.
\end{itemize}