Find all positive integers $n$, for which there exist positive integers $a>b$, satisfying $n=\frac{4ab}{a-b}$. 
1. We start with the given equation:
   \[
   n = \frac{4ab}{a-b}
   \]
   Define \( c = a - b \), where \( c > 0 \). Then we can rewrite the equation as:
   \[
   n = 4b + \frac{4b^2}{c}
   \]
   This implies that \( n > 4 \).

2. We need to find conditions under which \( n \) is an integer. For \( n \) to be an integer, \( \frac{4b^2}{c} \) must also be an integer. This means \( c \) must divide \( 4b^2 \).

3. Consider different values of \( c \):
   - If \( c = 4b^2 \), then:
     \[
     n = 4b + \frac{4b^2}{4b^2} = 4b + 1
     \]
   - If \( c = 2b^2 \), then:
     \[
     n = 4b + \frac{4b^2}{2b^2} = 4b + 2
     \]
   - If \( c = b^2 \), then:
     \[
     n = 4b + \frac{4b^2}{b^2} = 4b + 4
     \]

4. From the above, we see that \( n \) can take values of the form \( 4b + k \) where \( k \) is an integer. This implies that \( n \) can be any integer greater than 4 that is not congruent to 3 modulo 4.

5. Now, consider the case where \( n \) is a prime number \( p \equiv 3 \pmod{4} \). We have:
   \[
   cp = 4b(b + c)
   \]
   Since \( p \) is a prime, it cannot divide 4. Therefore, \( p \) must divide \( b + c \). Let \( c = 4bk \), then:
   \[
   kp = b(1 + 4k)
   \]
   Since \( p \nmid b \), it must be that \( p \mid 1 + 4k \). This implies:
   \[
   1 + 4k \equiv 0 \pmod{p} \implies 1 + 4k = mp \text{ for some integer } m
   \]
   This contradicts \( p \equiv 3 \pmod{4} \), as \( 1 + 4k \) would be congruent to 1 modulo 4, not 3.

6. Therefore, \( n \) cannot be a prime number congruent to 3 modulo 4.

Conclusion:
The numbers \( n > 4 \) that are not primes congruent to 3 modulo 4 work.

The final answer is \( \boxed{ n > 4 } \) and \( n \not\equiv 3 \pmod{4} \) or \( n \) is not a prime congruent to 3 modulo 4.