Determine all values of $n$ for which there is a set $S$ with $n$ points, with no 3 collinear, with the following property: it is possible to paint all points of $S$ in such a way that all angles determined by three points in $S$, all of the same color or of three different colors, aren't obtuse. The number of colors available is unlimited.
1. **Initial Observation**:
   - We need to determine the values of \( n \) for which there exists a set \( S \) with \( n \) points, no three of which are collinear, such that all angles determined by three points in \( S \) (either all of the same color or all of different colors) are not obtuse.
   - The number of colors available is unlimited.

2. **Five Points Analysis**:
   - Consider any five points in the plane with no three collinear. By the Erdős–Szekeres theorem, any set of five points in general position (no three collinear) must form a convex quadrilateral with one point inside it, which guarantees the existence of an obtuse angle.
   - Therefore, it is impossible to have five points all of the same color or all of different colors without determining an obtuse angle.

3. **Color Limitation**:
   - Since five points must determine an obtuse angle, we cannot use five or more colors. Each color can appear at most four times.
   - If we use four colors, each color can appear at most four times. However, we need to check if this configuration avoids obtuse angles.

4. **Four Points Configuration**:
   - Given four points with no three collinear, no three of them determine an obtuse angle only if they form a rectangle.
   - Suppose we use four colors, and some color appears twice. Let \( A, B, C, D \) be four points with different colors, and \( D' \) have the same color as \( D \). Then \( ABCD \) and \( ABCD' \) are both rectangles, which is impossible because \( D \) and \( D' \) are distinct.
   - Therefore, if we use four colors, we can only use each color once.

5. **Three Colors Configuration**:
   - With three colors, we can use each color four times. We can arrange the points as follows:
     - For each color, color four vertices of a small square that color.
     - Arrange the three squares near the vertices of a large equilateral triangle.
     - Rotate the squares so no collinearities occur, if necessary.

6. **Conclusion**:
   - For \( n \leq 12 \), we can arrange the points such that no three points of the same color or all different colors determine an obtuse angle.
   - For \( n > 12 \), it is impossible to avoid obtuse angles due to the limitations discussed.

The final answer is \( \boxed{ n \leq 12 } \).