How many pairs $(a, b)$ of positive integers $a,b$ solutions of the equation $(4a-b)(4b-a )=1770^n$  exist , if $n$ is a positive integer?
1. **Factorization of 1770:**
   \[
   1770 = 2 \cdot 3 \cdot 5 \cdot 59
   \]
   We denote \( A = 4a - b \) and \( B = 4b - a \). Given \( AB = 1770^n \), we need to find pairs \((a, b)\) such that \( A \) and \( B \) are positive integers.

2. **Divisibility Conditions:**
   Since \( AB = 1770^n \), both \( A \) and \( B \) must be divisible by the prime factors of 1770. Specifically:
   \[
   3 \mid AB \implies 3 \mid A \text{ or } 3 \mid B
   \]
   \[
   5 \mid AB \implies 5 \mid A \text{ or } 5 \mid B
   \]
   \[
   59 \mid AB \implies 59 \mid A \text{ or } 59 \mid B
   \]

3. **Further Divisibility Analysis:**
   Since \( 3 \mid AB \) and \( 5 \mid AB \), we can conclude:
   \[
   3 \mid A \text{ and } 3 \mid B
   \]
   \[
   5 \mid A \text{ and } 5 \mid B
   \]
   \[
   59 \mid A \text{ and } 59 \mid B
   \]
   Therefore, \( 15 \mid A \) and \( 15 \mid B \).

4. **Substitution:**
   Let \( A = 15c \) and \( B = 15d \) where \( c, d \in \mathbb{N} \). The system of equations becomes:
   \[
   4a - b = 15c
   \]
   \[
   4b - a = 15d
   \]
   Solving for \( a \) and \( b \):
   \[
   a = 4c + d
   \]
   \[
   b = c + 4d
   \]

5. **Case Analysis:**
   - **Case 1: \( n = 1 \)**
     \[
     15cd = 2 \cdot 59
     \]
     This equation has no integer solutions for \( c \) and \( d \).

   - **Case 2: \( n \geq 2 \)**
     \[
     15cd = 1770^n \implies cd = \frac{1770^n}{15} = 3^{n-2} \cdot 5^{n-2} \cdot 2^2 \cdot 59^2
     \]
     The number of natural divisors of \( 3^{n-2} \cdot 5^{n-2} \cdot 2^2 \cdot 59^2 \) is:
     \[
     (n-1) \cdot (n-1) \cdot 3 \cdot 3 = 9(n-1)^2
     \]
     Therefore, there are \( 9(n-1)^2 \) pairs \((c, d) \in \mathbb{N} \times \mathbb{N}\) such that \( cd = 3^{n-2} \cdot 5^{n-2} \cdot 2^2 \cdot 59^2 \).

6. **Conclusion:**
   The number of ordered pairs \((a, b)\) is equal to the number of ordered pairs \((c, d)\), which is \( 9(n-1)^2 \).

   - If \((a, b)\) are considered as ordered pairs, the number of solutions is:
     \[
     N_n = 9(n-1)^2
     \]

   - If \((a, b)\) are considered as sets of two numbers with \((a, b) = (b, a)\), the number of solutions is:
     \[
     M_n = \begin{cases} 
     \dfrac{9(n-1)^2}{2}, & \text{for } n \text{ odd} \\
     \dfrac{9(n-1)^2 + 1}{2}, & \text{for } n \text{ even}
     \end{cases}
     \]

The final answer is \( \boxed{ N_n = 9(n-1)^2 } \) for ordered pairs and \( M_n = \begin{cases} \dfrac{9(n-1)^2}{2}, & \text{for } n \text{ odd} \\ \dfrac{9(n-1)^2 + 1}{2}, & \text{for } n \text{ even} \end{cases} \) for sets.