For which integers $n \geq 4$ is the following procedure possible? Remove one number of the integers $1,2,3,\dots,n+1$ and arrange them in a sequence $a_1,a_2,\dots,a_n$ such that of the $n$ numbers \[ |a_1-a_2|,|a_2-a_3|,\dots,|a_{n-1}-a_n|,|a_n-a_1| \] no two are equal.
1. **Understanding the Problem:**
   We need to determine for which integers \( n \geq 4 \) it is possible to remove one number from the set \(\{1, 2, 3, \ldots, n+1\}\) and arrange the remaining \( n \) numbers in a sequence \( a_1, a_2, \ldots, a_n \) such that the absolute differences \(|a_1 - a_2|, |a_2 - a_3|, \ldots, |a_{n-1} - a_n|, |a_n - a_1|\) are all distinct.

2. **Sum of Differences:**
   The sum of the absolute differences is given by:
   \[
   |a_1 - a_2| + |a_2 - a_3| + \cdots + |a_{n-1} - a_n| + |a_n - a_1|
   \]
   Since the sequence is a permutation of \(\{1, 2, \ldots, n+1\}\) with one number removed, the differences range from 1 to \( n \). Therefore, the sum of these differences is:
   \[
   \sum_{k=1}^n k = \frac{n(n+1)}{2}
   \]

3. **Modulo 2 Analysis:**
   We need to compare the sum modulo 2. The sum \(\frac{n(n+1)}{2}\) must be an integer, which it always is for any integer \( n \). However, we need to ensure that the differences are distinct and cover all integers from 1 to \( n \).

4. **Constructing Sequences:**
   - For \( n = 4k - 1 \) with \( k \geq 2 \):
     \[
     (a_1, a_2, \ldots, a_{4k-1}) = (1, 4k, 2, 4k-1, \ldots, k, 3k+1, k+2, 3k, k+3, \ldots, 2k+2, 2k+1)
     \]
     This sequence can be formally defined as:
     \[
     a_{2i} = 4k + 1 - i \quad \text{for } 1 \leq i \leq 2k-1, \quad a_{2i-1} = \begin{cases} 
     i & \text{for } 1 \leq i \leq k, \\
     i + 1 & \text{for } k < i \leq 2k 
     \end{cases}
     \]
     This sequence satisfies the condition that all differences are distinct.

   - For \( n = 4 \):
     \[
     (a_1, a_2, a_3, a_4) = (1, 3, 2, 5)
     \]
     This sequence also satisfies the condition.

   - For \( n = 4k \) with \( k \geq 2 \):
     \[
     (a_1, a_2, \ldots, a_{4k-1}, a_{4k}) = (1, 4k, 2, 4k-1, \ldots, k, 3k+1, k+2, 3k, k+3, \ldots, 2k+2, 2k+1, 4k+1)
     \]
     This sequence can be verified to satisfy the condition.

5. **Conclusion:**
   We have shown that the procedure is possible for \( n \equiv 0 \pmod 4 \) and \( n \equiv 3 \pmod 4 \).

The final answer is \( \boxed{ n \equiv 0 \pmod 4 } \) or \( n \equiv 3 \pmod 4 \).