Numbers $1, 2, \dots , n$ are written around a circle in some order. What is the smallest possible sum of the absolute differences of adjacent numbers?
1. **Claim**: The smallest possible sum of the absolute differences of adjacent numbers is \(2n - 2\).

2. **Achievability**: 
   - Arrange the numbers \(1, 2, \ldots, n\) in increasing order around the circle.
   - The absolute differences between adjacent numbers are:
     \[
     |2 - 1| + |3 - 2| + \cdots + |n - (n-1)| + |1 - n|
     \]
   - This simplifies to:
     \[
     1 + 1 + \cdots + 1 + (n - 1) = (n - 1) + (n - 1) = 2n - 2
     \]
   - Therefore, \(2n - 2\) is achievable.

3. **Minimality**:
   - Consider the numbers \(1\) and \(n\) on the circle.
   - By the Triangle Inequality, the sum of the absolute differences along the minor arc between \(1\) and \(n\) must be at least \(n - 1\).
   - Similarly, the sum of the absolute differences along the other arc must also be at least \(n - 1\).
   - Therefore, the total sum of the absolute differences is at least:
     \[
     (n - 1) + (n - 1) = 2n - 2
     \]
   - Hence, \(2n - 2\) is the minimal possible sum.

\(\blacksquare\)

The final answer is \( \boxed{ 2n - 2 } \)