Le $S$ be the set of positive integers greater than or equal to $2$. A function $f: S\rightarrow S$ is italian if $f$ satifies all the following three conditions:
1) $f$ is surjective 
2) $f$ is increasing in the prime numbers(that is, if $p_1<p_2$ are prime numbers, then $f(p_1)<f(p_2)$)
3) For every $n\in S$ the number $f(n)$ is the product of $f(p)$, where $p$ varies among all the primes which divide $n$ (For instance, $f(360)=f(2^3\cdot 3^2\cdot 5)=f(2)\cdot f(3)\cdot f(5)$).
Determine the maximum and the minimum possible value of $f(2020)$, when $f$ varies among all italian functions.
To determine the maximum and minimum possible values of \( f(2020) \) for an Italian function \( f \), we need to carefully analyze the given conditions and properties of \( f \).

1. **Understanding the function \( f \)**:
   - \( f \) is surjective, meaning every positive integer greater than or equal to 2 is in the range of \( f \).
   - \( f \) is increasing on prime numbers, i.e., if \( p_1 < p_2 \) are primes, then \( f(p_1) < f(p_2) \).
   - For any \( n \in S \), \( f(n) \) is the product of \( f(p) \) for all primes \( p \) dividing \( n \).

2. **Prime factorization of 2020**:
   - \( 2020 = 2^2 \cdot 5 \cdot 101 \).
   - Therefore, \( f(2020) = f(2^2 \cdot 5 \cdot 101) = f(2)^2 \cdot f(5) \cdot f(101) \).

3. **Minimum value of \( f(2020) \)**:
   - To minimize \( f(2020) \), we need to minimize the values of \( f(2) \), \( f(5) \), and \( f(101) \).
   - Since \( f \) is increasing on primes and surjective, the smallest values \( f \) can take on primes are the smallest positive integers greater than or equal to 2.
   - Assign \( f(2) = 2 \), \( f(3) = 3 \), \( f(5) = 4 \), \( f(7) = 5 \), and so on.
   - For \( f(101) \), since 101 is the 26th prime number, \( f(101) \) should be at least 27.
   - Therefore, \( f(2020) \geq 2^2 \cdot 4 \cdot 27 = 4 \cdot 4 \cdot 27 = 432 \).

4. **Maximum value of \( f(2020) \)**:
   - To maximize \( f(2020) \), we need to maximize the values of \( f(2) \), \( f(5) \), and \( f(101) \).
   - Since \( f \) is surjective and increasing on primes, we can assign larger values to \( f(2) \), \( f(5) \), and \( f(101) \) while maintaining the increasing property.
   - Let \( f(2) = 2 \), \( f(3) = 3 \), \( f(5) = 5 \), \( f(7) = 7 \), and so on.
   - For \( f(101) \), since 101 is the 26th prime number, \( f(101) \) should be at least 101.
   - Therefore, \( f(2020) \leq 2^2 \cdot 5 \cdot 101 = 4 \cdot 5 \cdot 101 = 2020 \).

Conclusion:
- The minimum possible value of \( f(2020) \) is 432.
- The maximum possible value of \( f(2020) \) is 2020.

The final answer is \( \boxed{ 432 \leq f(2020) \leq 2020 } \).