Given an $n \times n \times n$ grid of unit cubes, a cube is [i]good[/i] if it is a sub-cube of the grid and has side length at least two. If a good cube contains another good cube and their faces do not intersect, the first good cube is said to [i]properly[/i] contain the second. What is the size of the largest possible set of good cubes such that no cube in the set properly contains another cube in the set?
1. **Define the problem and the concept of good cubes:**
   - A good cube is a sub-cube of an \( n \times n \times n \) grid with a side length of at least 2.
   - A good cube properly contains another good cube if the second cube is entirely within the first and their faces do not intersect.

2. **Define the progeny of a cube:**
   - The progeny of a cube is the largest cube properly contained within it.

3. **Consider the largest possible set of good cubes such that no cube properly contains another:**
   - Start with any maximal set of good cubes where no cube properly contains another.
   - Identify the largest cube in this set. If it is small (side length 2 or 3), we are done.
   - If it is large (side length 4 or greater), remove this cube and add its progeny.

4. **Iterate the process:**
   - The progeny is still a good cube and is properly contained within the removed cube, so it could not have been in the original set.
   - If any cube in the original set properly contains the progeny, it must be larger than the removed cube, which is a contradiction.
   - If the progeny properly contains any cube, the removed cube also properly contained it, which is another contradiction.
   - This process generates sets of cubes of the same order, fulfilling the pairwise-non-properly-containing property, with successively smaller total side lengths.

5. **Conclude the process:**
   - Eventually, this process stops, generating a set with no large cubes.
   - The set has an order no larger than the set of all \( 2 \times 2 \times 2 \) and \( 3 \times 3 \times 3 \) cubes.

6. **Calculate the maximal set:**
   - The number of \( 2 \times 2 \times 2 \) cubes in an \( n \times n \times n \) grid is \( (n-1)^3 \).
   - The number of \( 3 \times 3 \times 3 \) cubes in an \( n \times n \times n \) grid is \( (n-2)^3 \).
   - Therefore, the maximal set of good cubes is \( (n-1)^3 + (n-2)^3 \).

The final answer is \( \boxed{ (n-1)^3 + (n-2)^3 } \).