What is the smallest integer $n$ for which any subset of $\{1,2,3,\ldots,20\}$ of size $n$ must contain two numbers that differ by $8$?

$\textbf{(A) }2\qquad\textbf{(B) }8\qquad\textbf{(C) }12\qquad\textbf{(D) }13\qquad\textbf{(E) }15$
1. **Identify the problem constraints**: We need to find the smallest integer \( n \) such that any subset of \(\{1, 2, 3, \ldots, 20\}\) of size \( n \) must contain two numbers that differ by 8.

2. **Consider the Pigeonhole Principle**: The Pigeonhole Principle states that if \( n \) items are put into \( m \) containers, with \( n > m \), then at least one container must contain more than one item. Here, we can think of the numbers that differ by 8 as being in the same "container."

3. **Partition the set into groups**: We can partition the set \(\{1, 2, 3, \ldots, 20\}\) into groups where each group contains numbers that differ by 8:
   - Group 1: \(\{1, 9, 17\}\)
   - Group 2: \(\{2, 10, 18\}\)
   - Group 3: \(\{3, 11, 19\}\)
   - Group 4: \(\{4, 12, 20\}\)
   - Group 5: \(\{5, 13\}\)
   - Group 6: \(\{6, 14\}\)
   - Group 7: \(\{7, 15\}\)
   - Group 8: \(\{8, 16\}\)

4. **Determine the maximum size of a subset without two numbers differing by 8**: Each group can contribute at most one number to the subset without violating the condition. Since there are 8 groups, the maximum size of such a subset is 8.

5. **Calculate the smallest \( n \)**: If we have a subset of size 9, by the Pigeonhole Principle, at least one group must contain two numbers. Therefore, the smallest \( n \) for which any subset must contain two numbers that differ by 8 is \( 8 + 1 = 9 \).

6. **Verify the answer**: We need to ensure that for \( n = 9 \), any subset of size 9 from \(\{1, 2, 3, \ldots, 20\}\) will indeed contain two numbers that differ by 8. Since there are only 8 groups, selecting 9 numbers will force at least one group to have two numbers, confirming our solution.

The final answer is \(\boxed{9}\).