Given is natural number $n$. Sasha claims that for any $n$ rays in space, no two of which have a common point, he will be able to mark on these rays $k$ points lying on one sphere. What is the largest $k$ for which his statement is true? 
**
   - Consider a plane in space on which none of the rays lie.
   - Classify the rays based on whether they point upward or downward with respect to this plane.
   - By the pigeonhole principle, at least \( \lceil n/2 \rceil \) rays will point in the same direction (either upward or downward). Without loss of generality, assume they point upward.

3. **Intersection with a Plane:**
   - Since these \( \lceil n/2 \rceil \) rays all point upward, there exists a plane that intersects all of them.
   - Consider a sphere that almost degenerates into this plane, meaning it has a very large radius such that its part which tends to infinity is upwards.
   - This sphere will intersect each of the \( \lceil n/2 \rceil \) rays twice (once when the ray enters the sphere and once when it exits).

4. **Counting the Points:**
   - Therefore, the total number of intersection points on the sphere is \( 2 \times \lceil n/2 \rceil \).

5. **Proving the Upper Bound:**
   - To show that \( 2 \lceil n/2 \rceil \) is the maximum possible, consider the following construction:
     - Take \( \lceil n/2 \rceil \) lines.
     - On \( \lfloor n/2 \rfloor \) of these lines, place two rays with opposite directions.
     - On the remaining line (if \( n \) is odd), place a single ray.
   - Each line can have at most two intersections with the sphere.
   - Therefore, the total number of intersection points is:
     \[
     2 \times \lfloor n/2 \rfloor + 2 \times (\lceil n/2 \rceil - \lfloor n/2 \rfloor) = 2 \lceil n/2 \rceil
     \]

6. **Conclusion:**
   - Thus, the largest \( k \) for which Sasha's statement is true is \( 2 \lceil n/2 \rceil \).

The final answer is \( \boxed{ 2 \lceil n/2 \rceil } \).