A list of integers with average $89$ is split into two disjoint groups.  The average of the integers in the first group is $73$ while the average of the integers in the second group is $111$.  What is the smallest possible number of integers in the original list?
[i] Proposed by David Altizio [/i]
1. Let the total number of integers in the original list be \( n \). Let \( a \) be the number of integers in the first group and \( b \) be the number of integers in the second group. Therefore, we have \( a + b = n \).

2. Given that the average of the integers in the original list is 89, we can write the total sum of the integers in the original list as \( 89n \).

3. The average of the integers in the first group is 73, so the total sum of the integers in the first group is \( 73a \).

4. The average of the integers in the second group is 111, so the total sum of the integers in the second group is \( 111b \).

5. Since the total sum of the integers in the original list is the sum of the integers in both groups, we have:
   \[
   89n = 73a + 111b
   \]

6. Substituting \( n = a + b \) into the equation, we get:
   \[
   89(a + b) = 73a + 111b
   \]

7. Simplifying the equation, we get:
   \[
   89a + 89b = 73a + 111b
   \]

8. Rearranging the terms, we get:
   \[
   89a - 73a = 111b - 89b
   \]

9. Simplifying further, we get:
   \[
   16a = 22b
   \]

10. Dividing both sides by 2, we get:
    \[
    8a = 11b
    \]

11. This implies that \( a \) and \( b \) must be in the ratio \( 11:8 \). Therefore, the smallest possible values for \( a \) and \( b \) that satisfy this ratio are \( a = 11 \) and \( b = 8 \).

12. Thus, the smallest possible number of integers in the original list is:
    \[
    n = a + b = 11 + 8 = 19
    \]

Conclusion:
\[
\boxed{19}
\]