Find $ \#\left\{ (x,y)\in\mathbb{N}^2\bigg| \frac{1}{\sqrt{x}} -\frac{1}{\sqrt{y}} =\frac{1}{2016}\right\} , $ where $ \# A $ is the cardinal of $ A . $
1. **Substitution and Simplification:**
   We start by substituting \( x = d \xi \) and \( y = d \eta \) where \( (\xi, \eta) = 1 \). This implies that \( \xi \) and \( \eta \) are coprime. The given equation is:
   \[
   \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{y}} = \frac{1}{2016}
   \]
   Substituting \( x = d \xi \) and \( y = d \eta \), we get:
   \[
   \frac{1}{\sqrt{d \xi}} - \frac{1}{\sqrt{d \eta}} = \frac{1}{2016}
   \]
   Simplifying, we have:
   \[
   \frac{1}{\sqrt{d}} \left( \frac{1}{\sqrt{\xi}} - \frac{1}{\sqrt{\eta}} \right) = \frac{1}{2016}
   \]

2. **Rationality and Squaring:**
   Since \( \frac{1}{\sqrt{\xi}} - \frac{1}{\sqrt{\eta}} \) is rational, squaring both sides gives:
   \[
   \left( \frac{1}{\sqrt{\xi}} - \frac{1}{\sqrt{\eta}} \right)^2 = \left( \frac{1}{2016 \sqrt{d}} \right)^2
   \]
   This implies that \( \sqrt{\xi \eta} \) must be rational, hence \( \xi \eta \) is a perfect square. Therefore, \( \xi \) and \( \eta \) must be squares of integers. Let \( \xi = k^2 \) and \( \eta = l^2 \) where \( k, l \geq 1 \) and \( (k, l) = 1 \).

3. **Further Simplification:**
   Substituting \( \xi = k^2 \) and \( \eta = l^2 \) into the equation, we get:
   \[
   \frac{1}{\sqrt{d}} \left( \frac{1}{k} - \frac{1}{l} \right) = \frac{1}{2016}
   \]
   Simplifying further, we have:
   \[
   \frac{1}{\sqrt{d}} \cdot \frac{l - k}{kl} = \frac{1}{2016}
   \]
   Hence, \( d \) must also be a perfect square, say \( d = m^2 \). Thus, we have:
   \[
   \frac{1}{m} \cdot \frac{l - k}{kl} = \frac{1}{2016}
   \]

4. **Finding \( m \):**
   The fraction \( \frac{l - k}{kl} \) is irreducible, so \( l - k \) must divide \( m \). Let \( m = (l - k) \mu \). Substituting this into the equation, we get:
   \[
   \frac{1}{(l - k) \mu} \cdot \frac{l - k}{kl} = \frac{1}{2016}
   \]
   Simplifying, we obtain:
   \[
   \frac{1}{\mu} \cdot \frac{1}{kl} = \frac{1}{2016}
   \]
   Therefore, we have:
   \[
   kl \mu = 2016
   \]

5. **Counting Solutions:**
   We need to find the number of triples \( (k, l, \mu) \) such that \( (k, l) = 1 \) and \( kl \mu = 2016 \). The prime factorization of 2016 is:
   \[
   2016 = 2^5 \cdot 3^2 \cdot 7
   \]
   We need to distribute the factors of 2016 between \( k \) and \( l \) such that \( k \) and \( l \) are coprime.

6. **Distribution of Factors:**
   - For the factor \( 2^5 \), we have 6 choices (0 to 5) for the exponent in \( k \) or \( l \), but since \( k \) and \( l \) must be coprime, we have \( 1 + 2 \cdot 5 = 11 \) possibilities.
   - For the factor \( 3^2 \), we have 3 choices (0 to 2) for the exponent in \( k \) or \( l \), giving \( 1 + 2 \cdot 2 = 5 \) possibilities.
   - For the factor \( 7 \), we have 2 choices (0 or 1) for the exponent in \( k \) or \( l \), giving \( 1 + 2 \cdot 1 = 3 \) possibilities.

7. **Total Number of Solutions:**
   Multiplying the number of possibilities, we get:
   \[
   11 \cdot 5 \cdot 3 = 165
   \]

The final answer is \(\boxed{165}\).