(Solution of Yousif Alkhalawi, IMO 2025 team's candidate)
Since each cube $2 \times 2 \times 2$, there are at least 4 orange unit cubes (here we call it by square) so we want to minimize the number of orange squares.

Construction: Let number the layer $n \times n$ of cubes from $1 \to n$ (bottom to top). For each layer, consider the coordinate of a single squares as $(x, y)$ for $1 \le x, y \le n$.
* For the odd layers, we color squares at $(x, y)$ orange if $x, y$ are both even. The number of orange squares is $n^2$.
* For the even layers, we color square at $(x, y)$ orange if $xy$ is even. The number of orange squares is $(2n+1)^2 - (n+1)^2 = n(3n+2)$.

SAUDI ARABIAN IMO Booklet 2025
---
## Saudi Booklet 2025 — Page 22
22
Solution of Preselection tests

The cube $2 \times 2 \times 2$ will have 1 orange square in odd layer and 3 orange squares in even layers so it always have 4 orange squares as desired. The total orange squares is $(n+1)n^2 + n^2(3n+2) = n^2(4n+3)$. Thus the number of green squares is
$$(2n + 1)^3 - n^2(4n + 3) = 4n^3 + 9n^2 + 6n + 1.$$

We will prove this is the maximum number of green unit cubes by induction on $n$. For the base case $n=1$, in the cube of size 3, consider some 3 sub-cubes of size 2 at 3 opposite corners. The number of orange squares on these sub-cubes is at least $4 \times 3 = 12$. Note that there is at most 1 square appears in all of 3 sub-cubes, and there are at most 3 other squares appear in 2 of 3 sub-cubes. So the number of orange square in big-cube is at least $12 - 2 - 3 = 7$.

For a cube of size $2n+1$, we consider the sub-cube of size $2n-1$ in the top corner, denote it as $\Omega$ and the square in the opposite corner with it as $X$. By induction, $\Omega$ need at least $4(n-1)^3 + 3(n-1)^3$ orange squares so we need to prove in the rest of the big-cube, there are at least $(4n^3+3n^2) - (4(n-1)^3 + 3(n-1)^2) = 12n^2 - 6n + 1$ orange squares. Note that to build up the original cubes, one can place $n^2$ sub-cubes of size 2 on each face of $\Omega$ and one sub-cubes of size 2 touching $X$. So there are $3n^2+1$ sub-cubes leads to the sum of orange square is at least $12n^2+4$. To estimate the duplication, for each of $2n$ squares on the edge containing $X$, it can be in 2 sub-cubes of size 2 and for square $X$, it can be in 3 so the minimum is
$$4(3n^2 + 1) - 3 \times 2n - 1 \times 3 = 12n^2 - 6n + 1.$$
This finishes the proof. $\square$