This is a nice and quite difficult variant of APMO 2004/3. Take a point $O,$ color it red. Let all points an odd number of planes away from $O$ be red, and all others blue. This fails only if three points $A,B,C$ have $d_{AB}+d_{BC}+d_{CA}$ being even, where $d$ is the number of planes separating two points. A plane only matters in the above sum $\pmod{2}$ if it separates $1$ or $3$ of the pairs $\{A, B \}; \{B, C\} ;\{C, A\}.$ The latter case is impossible. In the former case, the plane goes through one of $A,B,$ or $C$ and separates the other two. Claim: Take any two points $X,Y$ out of the other $2002$ points that aren't $A,B,C.$ Then exactly one of the following is true: $AXY$ separates $\{B, C\},$ $BXY$ separates $\{C, A\},$ $CXY$ separates $\{A, B\}.$ Proof: Let $XY$ intersect plane $AXY$ at point $P.$ Then it's easy to see (just draw it out) that no matter where $P$ lies on the plane, exactly one of the following is true: line $AP$ intersects segment $BC$ (equivalent to $AXY$ separates $\{B, C\}$), line $BP$ passes through segment $CA$ (equivalent to $BXY$ separates $\{C, A\}$), line $CP$ passes through segment $AB$ (equivalent to $CXY$ separates $\{A, B\}$). $\square$ There are $\binom{2002}{2}$ such pairs, an odd number. So $d_{AB}+d_{BC}+d_{CA}$ is odd, the end. $\blacksquare$