Let us have $6050$ points in the plane, no three collinear. Find the maximum number $k$ of non-overlapping triangles without common vertices in this plane. 
1. **Determine the maximum number of non-overlapping triangles:**

   Given 6050 points in the plane, no three of which are collinear, we need to find the maximum number \( k \) of non-overlapping triangles without common vertices. 

   Since each triangle requires 3 vertices, the maximum number of triangles we can form is:
   \[
   k = \left\lfloor \frac{6050}{3} \right\rfloor = 2016
   \]

2. **Prove that 2016 is the desired number:**

   We need to show that it is possible to form 2016 non-overlapping triangles without common vertices. 

   Consider labeling the points as \( 1, 2, 3, \ldots, 6050 \). We can form a set of 2016 triangles by grouping the points in sets of three:
   \[
   (1, 2, 3), (4, 5, 6), (7, 8, 9), \ldots, (6046, 6047, 6048)
   \]
   This set of triangles is *relatively good* because no triangles share a vertex.

3. **Prove that the triangles in the relatively good set do not overlap:**

   Let \( T \) be a relatively good set with the minimal area of overlapping between its triangles. We will show that the triangles in \( T \) do not overlap.

   Assume, for contradiction, that two triangles, say \( (1, 2, 3) \) and \( (4, 5, 6) \), overlap.

   **Case 1: The line segment \( (4, 5) \) does not intersect the triangle \( (1, 2, 3) \).**

   The closest point in the triangle \( (1, 2, 3) \) to the segment \( (4, 5) \) is one of its vertices. Without loss of generality, assume it is vertex 1. Replace the triangles \( (1, 2, 3) \) and \( (4, 5, 6) \) with \( (1, 4, 5) \) and \( (2, 3, 6) \). These new triangles do not overlap because vertex 1 is the unique farthest point in the triangle \( (1, 4, 5) \) from the segment \( (4, 5) \). This contradicts the choice of \( T \) as having minimal overlapping area.

   **Case 2: The line segment \( (4, 5) \) intersects the triangle \( (1, 2, 3) \).**

   In this case, the line segment \( (4, 5) \) separates the vertices of the triangle \( (1, 2, 3) \) into two sets, one containing exactly one point and the other containing two points. Assume \( (4, 5) \) is horizontal and the lowest edge of the triangle \( (4, 5, 6) \).

   - If there is only one point below \( (4, 5) \), say vertex 1, then consider the triangles \( (1, 4, 5) \) and \( (2, 3, 6) \). These triangles do not intersect because we can separate them by taking a line parallel to \( (4, 5) \) that lies between \( (4, 5) \) and the points \( 2, 3, 6 \).

   - If there are two points below \( (4, 5) \), say vertices 1 and 2, then consider the triangles \( (1, 2, 4) \) and \( (5, 3, 6) \). One of these triangles is (almost) above the line \( (4, 5) \) and the other is below.

   In both cases, we can rearrange the triangles to avoid overlap, contradicting the assumption that the triangles in \( T \) overlap.

   Therefore, the triangles in \( T \) do not overlap, and the maximum number of non-overlapping triangles without common vertices is indeed 2016.

The final answer is \( \boxed{2016} \)