Let $A$ be a set of positive integers satisfying the following :

$a.)$ If $n \in A$ , then $n \le 2018$.

$b.)$ If $S \subset A$ such that $|S|=3$, then there exists $m,n \in S$ such that $|n-m| \ge \sqrt{n}+\sqrt{m}$

What is the maximum cardinality of $A$ ?
1. **Understanding the Problem:**
   We are given a set \( A \) of positive integers with two conditions:
   - \( a. \) If \( n \in A \), then \( n \leq 2018 \).
   - \( b. \) If \( S \subset A \) such that \( |S| = 3 \), then there exist \( m, n \in S \) such that \( |n - m| \geq \sqrt{n} + \sqrt{m} \).

2. **Rephrasing Condition \( b \):**
   We need to ensure that for any three elements in \( A \), there are at least two elements \( m \) and \( n \) such that:
   \[
   |n - m| \geq \sqrt{n} + \sqrt{m}
   \]
   This can be rewritten using the properties of square roots. For simplicity, let's assume \( n > m \). Then:
   \[
   n - m \geq \sqrt{n} + \sqrt{m}
   \]

3. **Analyzing the Inequality:**
   To understand the inequality better, consider the function \( f(x) = \sqrt{x} \). The difference between the square roots of two numbers \( n \) and \( m \) can be approximated by:
   \[
   \sqrt{n} - \sqrt{m} \approx \frac{n - m}{\sqrt{n} + \sqrt{m}}
   \]
   Therefore, the inequality \( n - m \geq \sqrt{n} + \sqrt{m} \) suggests that the elements of \( A \) must be spaced sufficiently apart.

4. **Spacing Elements:**
   To satisfy the inequality, we need to ensure that the elements of \( A \) are not too close to each other. Let's consider the spacing:
   \[
   n - m \geq \sqrt{n} + \sqrt{m}
   \]
   For simplicity, assume \( n \) and \( m \) are consecutive elements in \( A \). Then:
   \[
   n - m \geq \sqrt{n} + \sqrt{m}
   \]
   If we approximate \( n \approx m \), then:
   \[
   n - m \geq 2\sqrt{m}
   \]
   This suggests that the difference between consecutive elements should be at least \( 2\sqrt{m} \).

5. **Maximizing the Cardinality:**
   To maximize the cardinality of \( A \), we need to find the largest set of integers \( \{a_1, a_2, \ldots, a_k\} \) such that the difference between consecutive elements \( a_{i+1} - a_i \geq 2\sqrt{a_i} \).

6. **Constructing the Set:**
   Start with the smallest element, say \( a_1 = 1 \). The next element \( a_2 \) must satisfy:
   \[
   a_2 \geq 1 + 2\sqrt{1} = 3
   \]
   Continuing this process, we get:
   \[
   a_3 \geq 3 + 2\sqrt{3} \approx 3 + 3.46 \approx 6.46 \implies a_3 \geq 7
   \]
   \[
   a_4 \geq 7 + 2\sqrt{7} \approx 7 + 5.29 \approx 12.29 \implies a_4 \geq 13
   \]
   And so on.

7. **Finding the Maximum \( k \):**
   We need to find the largest \( k \) such that \( a_k \leq 2018 \). Using the approximation \( a_{i+1} \approx a_i + 2\sqrt{a_i} \), we can estimate the growth of \( a_i \).

8. **Estimating the Growth:**
   Using the approximation \( a_i \approx i^2 \), we find the largest \( i \) such that \( i^2 \leq 2018 \). Solving \( i^2 \leq 2018 \):
   \[
   i \leq \sqrt{2018} \approx 44.9
   \]
   Therefore, the maximum \( k \) is approximately 44.

The final answer is \( \boxed{44} \).