A figure on a computer screen shows $n$ points on a sphere, no four coplanar. Some pairs of points are joined by segments. Each segment is colored red or blue. For each point there is a key that switches the colors of all segments with that point as endpoint. For every three points there is a sequence of key presses that makes the three segments between them red. Show that it is possible to make all the segments on the screen red. Find the smallest number of key presses that can turn all the segments red, starting from the worst case.	

1. **Initial Setup and Definitions**:
   - Consider \( n \) points on a sphere, with no four points being coplanar.
   - Each pair of points is connected by a segment, which is either red or blue.
   - Each point has a key that switches the colors of all segments connected to it.
   - For any three points, there exists a sequence of key presses that makes all three segments between them red.

2. **Monochromatic Triangle**:
   - For any three points, it is possible to make all three segments red, forming a monochromatic red triangle.
   - This implies that the number of red segments in any triangle is odd.
   - The parity of the number of blue segments in a triangle does not change when switching keys, so the number of blue segments in each triangle is even.

3. **Partitioning Points**:
   - Let \( P \) be one of the \( n \) points.
   - Define \( A \) as the set of points connected to \( P \) by red segments.
   - Define \( B \) as the set of points connected to \( P \) by blue segments.

4. **Properties of Sets \( A \) and \( B \)**:
   - For any two points \( A_1, A_2 \in A \), the segments \( PA_1 \) and \( PA_2 \) are red, so \( A_1A_2 \) must be red.
   - For any two points \( B_1, B_2 \in B \), the segments \( PB_1 \) and \( PB_2 \) are blue, so \( B_1B_2 \) must be red.
   - For any point \( A \in A \) and \( B \in B \), the segment \( PA \) is red and \( PB \) is blue, so \( AB \) must be blue.

5. **Switching Keys**:
   - Place \( P \) in set \( A \).
   - Segments within the same set \( A \) or \( B \) are red, and segments connecting points in different sets \( A \) and \( B \) are blue.
   - Switching all points in set \( A \) will make all segments red:
     - Segments within \( A \) will change color twice (red to blue and back to red).
     - Segments between \( A \) and \( B \) will change color once (blue to red).
     - Segments within \( B \) will not change color.

6. **Minimizing Key Presses**:
   - Each point needs to be switched at most once.
   - Let \( |A| = k \) and \( |B| = n - k \).
   - If we switch \( a \) points from \( A \) and \( b \) points from \( B \), we change at most \( a(n-k) + bk \) blue segments.
   - Assume \( k \leq n - k \), hence \( k \leq \left\lfloor \frac{n}{2} \right\rfloor \).
   - Therefore, \( k(n-k) \leq a(n-k) + b(n-k) \), implying \( k \leq a + b \).
   - The number of key presses is at most \( k \), and in the worst case, \( \left\lfloor \frac{n}{2} \right\rfloor \).

The final answer is \( \boxed{ \left\lfloor \frac{n}{2} \right\rfloor } \)