Call a set of positive integers "conspiratorial" if no three of them are pairwise relatively prime. What is the largest number of elements in any "conspiratorial" subset of the integers $1$ to $16$?
1. **Identify the problem constraints**: We need to find the largest subset of the integers from 1 to 16 such that no three elements in the subset are pairwise relatively prime.

2. **Consider the set of prime numbers within the range**: The prime numbers between 1 and 16 are \( \{2, 3, 5, 7, 11, 13\} \). Any three of these numbers are pairwise relatively prime.

3. **Calculate the maximum size of the conspiratorial subset**: If a subset contains more than \(16 - 6 + 2 = 12\) elements, it must contain at least three of the prime numbers, which would be pairwise relatively prime. Therefore, the maximum size of a conspiratorial subset is \(11\).

4. **Construct a candidate subset**: Consider the set \( S = \{2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16\} \). We need to verify that no three elements in \( S \) are pairwise relatively prime.

5. **Verify the candidate subset**:
    - Every element in \( S \) is divisible by 2, 3, or 5.
    - Since 5 is not in \( S \), we only need to consider divisibility by 2 and 3.
    - The elements of \( S \) are: 
        - Divisible by 2: \( \{2, 4, 6, 8, 10, 12, 14, 16\} \)
        - Divisible by 3: \( \{3, 6, 9, 12, 15\} \)
    - Any three elements in \( S \) will share a common factor of either 2 or 3, ensuring they are not pairwise relatively prime.

6. **Conclusion**: Since \( S \) is a valid conspiratorial subset and has 11 elements, and we have shown that no larger conspiratorial subset can exist, we conclude that the largest conspiratorial subset of the integers from 1 to 16 has 11 elements.

The final answer is \( \boxed{ 11 } \).