Peter and Vasil together thought of ten 5-degree polynomials. Then, Vasil began calling consecutive natural numbers starting with some natural number. After each called number, Peter chose one of the ten polynomials at random and plugged in the called number. The results were recorded on the board. They eventually form a sequence. After they finished, their sequence was arithmetic. What is the greatest number of numbers that Vasil could have called out?

[i]A. Golovanov[/i]
1. Let \( P(x) \) be one of the ten 5-degree polynomials and let \( f(x) \) be the linear function modeling the arithmetic sequence. Since the sequence is arithmetic, \( f(x) \) can be written as:
   \[
   f(x) = ax + b
   \]
   where \( a \) and \( b \) are constants.

2. For the sequence to be arithmetic, the difference between consecutive terms must be constant. This implies that for each called number \( x \), the value \( P(x) \) must fit the form of \( f(x) \).

3. Consider the polynomial \( P(x) - f(x) \). Since \( P(x) \) is a 5-degree polynomial and \( f(x) \) is a linear polynomial, \( P(x) - f(x) \) is a 5-degree polynomial. 

4. A 5-degree polynomial can have at most 5 roots. Therefore, \( P(x) - f(x) = 0 \) can have at most 5 solutions for \( x \).

5. Since there are 10 different polynomials, each polynomial can intersect the line \( f(x) \) at most 5 times. Therefore, the total number of intersections (or solutions) is:
   \[
   10 \times 5 = 50
   \]

6. This means that the maximum number of distinct natural numbers that Vasil could have called out, such that the sequence remains arithmetic, is 50.

7. To illustrate this with an example, consider the polynomials \( P_k(x) = \prod_{i=1}^{5}(x - 5k - i) \) for \( k = 0, 1, \ldots, 9 \). The numbers called out could be \( 1, 2, \ldots, 50 \).

Conclusion:
The greatest number of numbers that Vasil could have called out is \( \boxed{50} \).