Find all integers $ n \ge 3$ such that there are $ n$ points in space, with no three on a line and no four on a circle, such that all the circles pass through three points between them are congruent.
To solve the problem, we need to find all integers \( n \ge 3 \) such that there are \( n \) points in space, with no three on a line and no four on a circle, such that all the circles passing through any three of these points are congruent.

1. **Case \( n = 4 \):**
   - Consider a regular tetrahedron. In a regular tetrahedron, any three vertices form an equilateral triangle, and the circumcircle of any such triangle is congruent to the circumcircle of any other triangle formed by three vertices of the tetrahedron. Therefore, for \( n = 4 \), the condition is satisfied.

2. **Case \( n \ge 5 \):**
   - **Subcase 1: Four points lie on a plane.**
     - Suppose there are four points \( A, B, C, D \) that lie on a plane. Since no three points are collinear and no four points are concyclic, the circles passing through any three of these points must be congruent.
     - Consider the circles passing through points \( A, B, C \) and \( A, B, D \). These circles must have the same radius. However, since there are only two circles with a fixed radius passing through two fixed points \( A \) and \( B \), and if at least one of them contains the other two points, it leads to a contradiction.
     - Therefore, there must be exactly one point on each circle, say \( C \) and \( D \). Applying the same argument with two fixed points \( B \) and \( C \) leads to another contradiction.
   
   - **Subcase 2: No four points lie on a plane.**
     - Consider any three vertices \( A, B, C \). There must be two other points \( D \) and \( E \) in space. We apply the following lemma:
       - **Lemma:** For any tetrahedron \( ABCD \) such that the circumradius of \( \triangle ABC \), \( \triangle ADC \), \( \triangle ABD \), and \( \triangle DBC \) are equal, then \( AB = CD \), \( AD = BD \), and \( AD = BC \).
     - Using this lemma, the distances between \( D \) and \( E \) to \( A, B, C \) are fixed. Therefore, \( D \) and \( E \) are uniquely determined from \( A, B, C \).
     - Applying the same argument with \( A, B, E \), where \( C \) and \( D \) lie on the same side of the plane passing through these three points, we get that \( D = C \), which is a contradiction.

From the above cases, we conclude that the only possible value for \( n \) is 4.

The final answer is \( \boxed{ n = 4 } \).