For what integers $ n\ge 3$ is it possible to accommodate, in some order, the numbers $ 1,2,\cdots, n$ in a circular form such that every number divides the sum of the next two numbers, in a clockwise direction?
1. **Understanding the Problem:**
   We need to arrange the numbers \(1, 2, \ldots, n\) in a circular form such that each number divides the sum of the next two numbers in a clockwise direction. We need to determine for which integers \(n \geq 3\) this is possible.

2. **Analyzing Parity:**
   - If two even numbers are adjacent, then all numbers must be even, which is not possible since \(n\) includes both odd and even numbers.
   - Therefore, an even number must be surrounded by odd numbers. This implies a pattern of odd-even-odd.

3. **Pattern Deduction:**
   - Consider the sequence: odd-odd-even-odd-odd. This pattern suggests that for every even number, there must be at least two odd numbers.
   - If \(n\) is even, say \(n = 2d\), then there are \(d\) odd numbers and \(d\) even numbers. This configuration is not possible because each even number requires two odd numbers around it, which is not feasible with equal numbers of odd and even numbers.

4. **Odd \(n\):**
   - If \(n\) is odd, say \(n = 2d + 1\), then there are \(d\) even numbers and \(d + 1\) odd numbers.
   - For the arrangement to work, the number of odd numbers must be at least twice the number of even numbers, i.e., \(d + 1 \geq 2d\).
   - Solving the inequality: \(d + 1 \geq 2d\) gives \(1 \geq d\), so \(d = 1\).

5. **Conclusion:**
   - If \(d = 1\), then \(n = 2d + 1 = 3\).
   - For \(n = 3\), the numbers are \(1, 2, 3\). We can arrange them in a circle as \(1, 2, 3\) such that:
     - \(1\) divides \(2 + 3 = 5\)
     - \(2\) divides \(3 + 1 = 4\)
     - \(3\) divides \(1 + 2 = 3\)

Thus, the only integer \(n \geq 3\) for which it is possible to arrange the numbers \(1, 2, \ldots, n\) in a circular form such that each number divides the sum of the next two numbers is \(n = 3\).

The final answer is \(\boxed{3}\).