A cubic block with dimensions $n$ by $n$ by $n$ is made up of a collection of $1$ by $1$ by $1$ unit cubes. What is the smallest value of $n$ so that if the outer two layers of unit cubes are removed from the block, more than half the original unit cubes will still remain?
To solve this problem, we need to determine the smallest value of \( n \) such that removing the outer two layers of unit cubes from an \( n \times n \times n \) cubic block leaves more than half of the original unit cubes.

1. **Calculate the total number of unit cubes in the original block:**
   \[
   \text{Total number of unit cubes} = n^3
   \]

2. **Determine the dimensions of the inner cube after removing the outer two layers:**
   - Removing two layers from each face reduces each dimension by 4 (2 layers from each side).
   - The dimensions of the inner cube are \((n-4) \times (n-4) \times (n-4)\).

3. **Calculate the number of unit cubes in the inner cube:**
   \[
   \text{Number of unit cubes in the inner cube} = (n-4)^3
   \]

4. **Set up the inequality to ensure that the number of unit cubes in the inner cube is more than half of the original number of unit cubes:**
   \[
   (n-4)^3 > \frac{n^3}{2}
   \]

5. **Solve the inequality:**
   \[
   (n-4)^3 > \frac{n^3}{2}
   \]
   Expanding the left-hand side:
   \[
   n^3 - 12n^2 + 48n - 64 > \frac{n^3}{2}
   \]
   Subtract \(\frac{n^3}{2}\) from both sides:
   \[
   n^3 - 12n^2 + 48n - 64 - \frac{n^3}{2} > 0
   \]
   Combine like terms:
   \[
   \frac{n^3}{2} - 12n^2 + 48n - 64 > 0
   \]
   Multiply through by 2 to clear the fraction:
   \[
   n^3 - 24n^2 + 96n - 128 > 0
   \]

6. **Find the smallest integer \( n \) that satisfies the inequality:**
   - By inspection or trial and error, we test values of \( n \):
     - For \( n = 20 \):
       \[
       20^3 - 24 \cdot 20^2 + 96 \cdot 20 - 128 = 8000 - 9600 + 1920 - 128 = 192 > 0
       \]
     - For \( n = 19 \):
       \[
       19^3 - 24 \cdot 19^2 + 96 \cdot 19 - 128 = 6859 - 8646 + 1824 - 128 = -91 < 0
       \]
   - Therefore, the smallest \( n \) that satisfies the inequality is \( n = 20 \).

The final answer is \( \boxed{20} \).