Place all numbers from 1 to 10 to the boxes such that every number except the uppermost is equal to the difference between the two numbers on its top. [asy]
unitsize(-4);
draw((0,0)--(5,0)--(5,5)--(0,5)--cycle);
draw((10,0)--(15,0)--(15,5)--(10,5)--cycle);
draw((20,0)--(25,0)--(25,5)--(20,5)--cycle);
draw((30,0)--(35,0)--(35,5)--(30,5)--cycle);

draw((5,10)--(10,10)--(10,15)--(5,15)--cycle);
draw((15,10)--(20,10)--(20,15)--(15,15)--cycle);
draw((25,10)--(30,10)--(30,15)--(25,15)--cycle);

draw((10,20)--(15,20)--(15,25)--(10,25)--cycle);
draw((20,20)--(25,20)--(25,25)--(20,25)--cycle);

draw((15,30)--(20,30)--(20,35)--(15,35)--cycle);

draw((2.5,5)--(7.5, 10));
draw((12.5,5)--(17.5, 10));
draw((22.5,5)--(27.5, 10));

draw((32.5,5)--(27.5, 10));
draw((22.5,5)--(17.5, 10));
draw((12.5,5)--(7.5, 10));

draw((7.5,15)--(12.5, 20));
draw((17.5,15)--(22.5, 20));

draw((27.5,15)--(22.5, 20));
draw((17.5,15)--(12.5, 20));

draw((12.5,25)--(17.5, 30));
draw((22.5,25)--(17.5, 30));
[/asy]The number in the lower box is at most

$\textbf{(A)}\ 1  \qquad\textbf{(B)}\ 2  \qquad\textbf{(C)}\ 3  \qquad\textbf{(D)}\ 4  \qquad\textbf{(E)}\ 5$
To solve this problem, we need to place the numbers from 1 to 10 in the boxes such that each number, except the topmost one, is the difference between the two numbers directly above it. We will start by analyzing the constraints and then systematically place the numbers.

1. **Identify the structure and constraints:**
   - The pyramid has 4 levels.
   - The topmost level has 1 box.
   - The second level has 2 boxes.
   - The third level has 3 boxes.
   - The bottom level has 4 boxes.
   - Each number in a box (except the topmost) is the absolute difference of the two numbers directly above it.

2. **Analyze possible placements:**
   - The topmost number must be the largest number because it is the sum of the two numbers directly below it.
   - The numbers must be placed such that the differences are valid and unique.

3. **Start with the largest number at the top:**
   - Place 10 at the topmost box.

4. **Determine the numbers for the second level:**
   - The numbers in the second level must be such that their absolute difference is 10.
   - Possible pairs are (9, 1) or (1, 9).

5. **Check the placement of (9, 1) and (1, 9):**
   - If we place 9 and 1 in the second level, the third level must have numbers whose differences are 9 and 1.
   - This leads to a contradiction because we cannot place the remaining numbers to satisfy the conditions.

6. **Try another approach:**
   - Place 9 and 1 in the second level.
   - The third level must have numbers whose differences are 8 and 1.
   - Possible pairs are (8, 1) and (1, 8).

7. **Check the placement of (8, 1) and (1, 8):**
   - If we place 8 and 1 in the third level, the bottom level must have numbers whose differences are 7, 1, and 1.
   - This leads to a contradiction because we cannot place the remaining numbers to satisfy the conditions.

8. **Try another approach:**
   - Place 8 and 2 in the second level.
   - The third level must have numbers whose differences are 6 and 2.
   - Possible pairs are (6, 2) and (2, 6).

9. **Check the placement of (6, 2) and (2, 6):**
   - If we place 6 and 2 in the third level, the bottom level must have numbers whose differences are 4, 2, and 2.
   - This leads to a contradiction because we cannot place the remaining numbers to satisfy the conditions.

10. **Try another approach:**
    - Place 7 and 3 in the second level.
    - The third level must have numbers whose differences are 4 and 3.
    - Possible pairs are (4, 3) and (3, 4).

11. **Check the placement of (4, 3) and (3, 4):**
    - If we place 4 and 3 in the third level, the bottom level must have numbers whose differences are 1, 3, and 1.
    - This leads to a contradiction because we cannot place the remaining numbers to satisfy the conditions.

12. **Conclusion:**
    - After trying all possible placements, we find that the smallest number in the bottom row is 1.

The final answer is $\boxed{1}$.