I found this problem in Chinese book of riddles. I'm sorry that I can't provide a reference for the book: it appears really old and the cover is blank.
Using white cubes that measure $1\times 1\times 1$ make a cube that measures $4\times 4\times 4$ and paint its surface red. Then rearrange the same $1 \times 1 \times 1$ cubes into another white cube and again paint its surface red. Is it necessarily true that one can still make a white $4\times 4 \times 4$ cube from these smaller cubes?
I divided the $1\times 1\times 1$ cubes in four different groups: (1)interior, (2)face, (3)edge, (4)corner. So after the first iteration these cubes have 96 red surfaces, and after the second iteration 192 red surfaces, which leaves sufficiently many white faces to make one more white cube (if they are located properly). But I guess there's one more improper way, which will lead us to the solution (that there's a case where we cannot still make a white cube)?
Call a piece a $C$-piece if it is a corner piece, $E$ for edge, $I$ for inside, and $F$ for face. We can force the pieces to land in the right spot so that the end result is a white cube.
Then for any $I$-piece, after painting the first time, and painting the second time, we can always make it into a $C$-piece. So, the path of an $I$-piece is from inside, to wherever, to a corner. Denote this as:
$$I \longrightarrow \star \longrightarrow C$$
The rest have similar properties, in fact,
$$\begin{align} I &\longrightarrow \star \longrightarrow C \\ F &\longrightarrow \star \longrightarrow E \\ C &\longrightarrow \star \longrightarrow I \\ E &\longrightarrow \star \longrightarrow F \\ \end{align} $$
Thus, given any painting, then rearrangement, and painting again, we can determine where the pieces should go to make a white cube.
Moreover, the cardinalities match up, so each map is bijection.