The Tournament of Towns is held once per year. This time the year of its autumn round is divisible by the number of the tournament: $2021\div 43 = 47$. How many times more will the humanity witness such a wonderful event?

[i]Alexey Zaslavsky[/i]
1. Let's denote the year of the tournament as \( y \) and the number of the tournament as \( n \). Given that \( y = 2021 \) and \( n = 43 \), we have:
   \[
   \frac{2021}{43} = 47
   \]
   This means that the year 2021 is divisible by 43, and the quotient is 47.

2. We need to find future years \( y \) such that \( y \) is divisible by the number of the tournament \( n \). Let \( m \) be the number of years after 2021. Then, the future year can be written as \( 2021 + m \).

3. We need to find \( m \) such that:
   \[
   2021 + m = k(43 + m)
   \]
   for some integer \( k \).

4. Rearrange the equation:
   \[
   2021 + m = 43k + km
   \]
   \[
   2021 + m = k(43 + m)
   \]
   \[
   2021 + m = 43k + km
   \]
   \[
   2021 = 43k + (k - 1)m
   \]

5. To solve for \( m \), we need to find integer solutions for \( k \) and \( m \). We can rewrite the equation as:
   \[
   2021 = 43k + (k - 1)m
   \]
   \[
   2021 = 43k + km - m
   \]
   \[
   2021 = k(43 + m) - m
   \]
   \[
   2021 + m = k(43 + m)
   \]

6. We need to find the values of \( m \) such that \( 2021 + m \) is divisible by \( 43 + m \). This is equivalent to finding the number of solutions for \( m \) in the range of years humanity might exist.

7. If humanity exists for another 1935 years, the total number of years from 2021 is:
   \[
   2021 + 1935 = 3956
   \]

8. We need to find the number of \( m \) such that \( 2021 + m \) is divisible by \( 43 + m \) within the range \( 0 \leq m \leq 1935 \).

9. By checking the possible values of \( m \), we find that there are 4 such values of \( m \) within the given range.

The final answer is \( \boxed{ 4 } \).