Find maximal numbers of planes, such there are $6$ points and 
1) $4$ or more  points lies on every plane.
2) No one line passes through $4$ points.
To solve the problem, we need to find the maximum number of planes such that:
1. Each plane contains at least 4 points.
2. No line passes through 4 points.

Let's analyze the problem step by step.

1. **Case 1: All 6 points lie on a single plane.**
   - If all 6 points lie on a single plane, then we can only have one plane. This is because any additional plane would be the same as the first one, and no line in that plane can contain 4 points. This case does not satisfy the conditions.

2. **Case 2: 5 points lie on a single plane.**
   - If 5 points lie on a single plane, then the remaining point must lie outside this plane. Any plane containing this outside point and any 3 of the 5 points in the plane will satisfy the conditions. However, this configuration does not maximize the number of planes, as we will see in the next cases.

3. **Case 3: No plane contains more than 4 points.**
   - Let's consider a plane \( H \) with exactly 4 points, say \( A, B, C, D \). The remaining 2 points, say \( P \) and \( Q \), lie outside this plane.
   - Any plane containing \( P \) and 3 of the 4 points in \( H \) will satisfy the conditions. Similarly, any plane containing \( Q \) and 3 of the 4 points in \( H \) will also satisfy the conditions.
   - We can form the following planes:
     - \( PABC \)
     - \( PABD \)
     - \( PBCD \)
     - \( QABC \)
     - \( QABD \)
     - \( QBCD \)
   - This gives us a total of 6 planes.

4. **Verification:**
   - Each of these planes contains exactly 4 points.
   - No line passes through 4 points, as each plane contains exactly 4 points and no more.

Therefore, the maximum number of planes that satisfy the given conditions is 6.

The final answer is \(\boxed{6}\)