Let $S$ be a set of $13$ distinct, pairwise relatively prime, positive integers. What is the smallest possible value of $\max_{s \in S} s- \min_{s \in S}s$?

[i]Proposed by James Lin
1. **Initial Consideration**: We need to find the smallest possible value of $\max_{s \in S} s - \min_{s \in S} s$ for a set $S$ of 13 distinct, pairwise relatively prime, positive integers. 

2. **Lower Bound**: Since the integers are pairwise relatively prime, they must be distinct and cannot share any common factors other than 1. The smallest 13 distinct positive integers are $1, 2, 3, \ldots, 13$. However, these are not pairwise relatively prime. We need to find a set of 13 such integers that are pairwise relatively prime and calculate the difference between the maximum and minimum values in this set.

3. **Odd Integers and Multiples**: Consider the set of odd integers. Among any string of 12 consecutive odd integers, at least 2 are multiples of 5. Thus, we need to add 2 to remove one of them. There are also at least 4 odd multiples of 3. To remove 3 of them, we must add 6. However, one can be a multiple of 15, so we can subtract 2 back. For multiples of 7, there is at least 1 of them. We don't need to remove this one. But note that there is at least 1 multiple of 21, so we must subtract 2 from our count.

4. **Tracking Odd Numbers**: Since we moved up 4 odd numbers total, we keep track of 12 + 4 = 16 odd numbers total. Within 16 odd numbers, at least 3 are multiples of 5, so we must add an additional 2. Similarly, there are at least 5 multiples of 3, so we must add another additional 2. Since there is now an additional odd multiple of 7, we must remove one of them to add 2.

5. **Final Calculation**: Summarizing the adjustments:
   \[
   22 + 2 + 6 - 2 - 2 + 2 + 2 + 2 = 32
   \]

6. **Example Set**: An example set that satisfies the conditions is $S = \{17, 19, 23, 25, 27, 29, 31, 32, 37, 41, 43, 47, 49\}$.

Conclusion:
\[
\boxed{32}
\]