Ten points are marked in the plane so that no three of them lie on a line. Each pair of points is connected with a segment. Each of these segments is painted with one of $k$ colors, in such a way that for any $k$ of the ten points, there are $k$ segments each joining two of them and no two being painted with the same color. Determine all integers $k$, $1\leq k\leq 10$, for which this is possible.
1. **Understanding the Problem:**
   We are given 10 points in the plane with no three points collinear. Each pair of points is connected by a segment, and each segment is painted with one of \( k \) colors. For any \( k \) of the 10 points, there must be \( k \) segments each joining two of them, with no two segments sharing the same color. We need to determine the possible values of \( k \) for which this is possible.

2. **Initial Observations:**
   - There are a total of \( \binom{10}{2} = 45 \) segments.
   - For \( k \leq 2 \), it is impossible to have \( k \) segments with \( k \) different colors among any \( k \) points because there are fewer than \( k \) segments.

3. **Case \( k = 3 \):**
   - Suppose \( k = 3 \). There are \( \binom{10}{3} = 120 \) sets of three points.
   - If each set of three points must have three segments with different colors, then each color must appear in at least one segment for each set of three points.
   - If a color \( C \) appears at most 15 times, then summing the number of edges \( C \) for all sets of three points, each edge is counted 8 times (since each edge is part of 8 different sets of three points).
   - This gives \( 15 \times 8 = 120 \), which matches the total number of sets of three points. However, this implies that each set of three points must contain exactly one edge of color \( C \), leading to a contradiction as explained in the solution.

4. **Case \( k = 4 \):**
   - Suppose \( k = 4 \). There are \( \binom{10}{4} = 210 \) sets of four points.
   - If each set of four points must have four segments with different colors, then each color must appear in at least one segment for each set of four points.
   - If a color \( C \) appears at most 11 times, then summing the number of edges \( C \) for all sets of four points, each edge is counted 6 times (since each edge is part of 6 different sets of four points).
   - This gives \( 11 \times 6 = 66 \), which is less than the total number of sets of four points, leading to a contradiction as explained in the solution.

5. **Case \( k \geq 5 \):**
   - For \( k \geq 5 \), we can create \( k-1 \) groups of vertices and color all edges between vertices of the same group.
   - For \( k = 5 \), we can use the following coloring scheme:
     \[
     \begin{align*}
     C_1 &: 012, 369, 48, 57 \\
     C_2 &: 234, 581, 60, 79 \\
     C_3 &: 456, 703, 82, 91 \\
     C_4 &: 678, 925, 04, 13 \\
     C_5 &: 890, 147, 26, 35
     \end{align*}
     \]
   - This ensures that for any set of 5 points, there will be at least one edge of each color among them.

6. **Generalization for \( k \geq 6 \):**
   - For \( k \geq 6 \), we can similarly ensure that the edges of the same color do not share a vertex, thus satisfying the condition for any \( k \) points.

The final answer is \( \boxed{ k \geq 5 } \).