Read the following information carefully and answer the questions given below:

Nine boxes - A, B, C, D, E, F, G, H and I are kept one above another in a single stack, but not necessarily in the same order. Only one box is kept between box E and box G. Box I is kept three boxes above box G. Box E is kept four boxes below box H. As many boxes are kept between box H and box I as between box D and box E. Box E is kept above box D. Box F is kept immediately below box C. Only two boxes are kept between box A and box F. Box B is kept above box A.

If all the boxes are arranged in alphabetical order from bottom to the top in alphabetical order, then how many boxes will not change their position?

To find the number of boxes that do not change their position when rearranged alphabetically, we first determine the step-by-step arrangement of the nine boxes (A, B, C, D, E, F, G, H, I) from bottom (1) to top (9) based on the clues provided.

Step 1: Analyze the clues regarding the positions of boxes G, I, E, H, and D.
1. Box I is kept three boxes above box G:
$Position(I)\; =\; Position(G)\; +\; 3$
2. Only one box is kept between box E and box G. This gives two possibilities: $Position(E)\; =\; Position(G)\; +\; 2$ or $Position(E)\; =\; Position(G)\; -\; 2$.
3. Box E is kept four boxes below box H:
$Position(H)\; =\; Position(E)\; +\; 4$
4. Box E is kept above box D ($Position(E)\; >\; Position(D)$), and the number of boxes between H and I is equal to the number of boxes between D and E.

Step 2: Testing the cases for E and G.
Let us test the case where $Position(E)\; =\; Position(G)\; +\; 2$.
Under this case:
$Position(H)\; =\; (Position(G)\; +\; 2)\; +\; 4\; =\; Position(G)\; +\; 6$
Let the position of G be $x$. Then the relative positions are:
- Box G: $x$
- Box E: $x\; +\; 2$
- Box I: $x\; +\; 3$
- Box H: $x\; +\; 6$
The number of boxes between H (at $x\; +\; 6$) and I (at $x\; +\; 3$) is 2 (at positions $x\; +\; 4$ and $x\; +\; 5$).
Therefore, there must be 2 boxes between D and E. Since E is above D, the position of D is:
$Position(D)\; =\; Position(E)\; -\; 3\; =\; x\; +\; 2\; -\; 3\; =\; x\; -\; 1$

Since there are 9 positions (1 to 9), the span of positions is from $x\; -\; 1$ (D) to $x\; +\; 6$ (H), which covers 8 boxes. Thus, $x$ can be either 2 or 3.
If we set $x\; =\; 2$ (meaning G is at position 2):
- Position 9: (empty)
- Position 8: H
- Position 7: (empty)
- Position 6: (empty)
- Position 5: I
- Position 4: E
- Position 3: (empty)
- Position 2: G
- Position 1: D

Step 3: Place the remaining boxes (A, B, C, F).
1. Box F is kept immediately below box C. This means C and F must occupy consecutive empty positions, with C directly above F.
Looking at the empty positions (9, 7, 6, 3), the only consecutive empty positions are 7 and 6. Therefore, we place C at 7 and F at 6.
2. Only two boxes are kept between box A and box F.
Since F is at position 6, A must be at position 3 (so that positions 4 and 5 are between them).
3. Box B is kept above box A.
With A at position 3, the only remaining empty position is 9, which is indeed above A. Thus, B is at position 9.

This completes the unique valid arrangement from bottom (1) to top (9):
- Position 9 (Top): B
- Position 8: H
- Position 7: C
- Position 6: F
- Position 5: I
- Position 4: E
- Position 3: A
- Position 2: G
- Position 1 (Bottom): D

Step 4: Arrange the boxes alphabetically from bottom to top and compare.
Let us compare the original arrangement with the alphabetical arrangement (A to I from bottom 1 to top 9):

Comparing the two sequences, we find that exactly two boxes, F (at position 6) and H (at position 8), do not change their positions.