Let $k \in \mathbb{N},$ $S_k = \{(a, b) | a, b = 1, 2, \ldots, k \}.$ Any two elements $(a, b)$, $(c, d)$ $\in S_k$ are called "undistinguishing" in $S_k$ if $a - c \equiv 0$ or $\pm 1 \pmod{k}$ and $b - d \equiv 0$ or $\pm 1 \pmod{k}$; otherwise, we call them "distinguishing". For example, $(1, 1)$ and $(2, 5)$ are undistinguishing in $S_5$. Considering the subset $A$ of $S_k$ such that the elements of $A$ are pairwise distinguishing. Let $r_k$ be the maximum possible number of elements of $A$.

(i) Find $r_5$.
(ii) Find $r_7$.
(iii) Find $r_k$ for $k \in \mathbb{N}$.
1. **Understanding the Problem:**
   - We are given a set \( S_k = \{(a, b) \mid a, b = 1, 2, \ldots, k\} \).
   - Two elements \((a, b)\) and \((c, d)\) are "undistinguishing" if \(a - c \equiv 0 \text{ or } \pm 1 \pmod{k}\) and \(b - d \equiv 0 \text{ or } \pm 1 \pmod{k}\).
   - We need to find the maximum subset \( A \subseteq S_k \) such that all elements in \( A \) are pairwise distinguishing.

2. **Initial Cases:**
   - For \( k = 1, 2, 3 \), it is straightforward to see that \( r_k = 1 \) because any two elements in \( S_k \) will be undistinguishing.

3. **General Case for \( k \geq 4 \):**
   - We define \( A_i = \{(i, b) \mid b = 1, 2, \ldots, k\} \) for \( i = 1, 2, \ldots, k \).
   - Define \( B_i = A_i \cup A_{i+1} \) for \( i = 1, 2, \ldots, k \) (with \( A_{k+1} = A_1 \)).

4. **Bounding \( |A \cap B_i| \):**
   - Consider \( B_1 \). The elements \((1, b)\), \((2, b)\), \((1, b+1)\), and \((2, b+1)\) are pairwise undistinguishing.
   - Therefore, the elements of \( A \) can have at most \( \lfloor \frac{k}{2} \rfloor \) different \( b \)-values, and for each \( b \), there is at most one \( a = 1 \) or \( 2 \).
   - Hence, \( |A \cap B_i| \leq \lfloor \frac{k}{2} \rfloor \).

5. **Summing Over All \( B_i \):**
   - If \( |A| = r_k \), then consider:
     \[
     \sum_{i=1}^k |A \cap B_i| = \sum_{i=1}^k (|A \cap A_i| + |A \cap A_{i+1}|) = 2 \sum_{i=1}^k |A \cap A_i| = 2|A| = 2r_k
     \]
   - Since \( |A \cap B_i| \leq \lfloor \frac{k}{2} \rfloor \), we have:
     \[
     2r_k \leq k \lfloor \frac{k}{2} \rfloor
     \]
   - Therefore, \( r_k \leq \left\lfloor \frac{k}{2} \left\lfloor \frac{k}{2} \right\rfloor \right\rfloor \).

6. **Constructing the Maximum Subset \( A \):**
   - We need to show that \( r_k = \left\lfloor \frac{k}{2} \left\lfloor \frac{k}{2} \right\rfloor \right\rfloor \).

   - **Case 1: \( k \) is even.**
     - Let \( k = 2m \). Then \( \left\lfloor \frac{k}{2} \left\lfloor \frac{k}{2} \right\rfloor \right\rfloor = m^2 \).
     - Define \( A = \{(a, b) \mid a, b \equiv 1 \pmod{2}, (a, b) \in S_k\} \).
     - It is easy to see that \( |A| = m^2 \) and any two elements \((a, b)\) and \((c, d)\) in \( A \) are distinguishing.

   - **Case 2: \( k \equiv 1 \pmod{4} \).**
     - Let \( k = 4t + 1 \). Then \( \left\lfloor \frac{k}{2} \left\lfloor \frac{k}{2} \right\rfloor \right\rfloor = t(4t + 1) \).
     - Define \( A \cap A_{2p+1} = \{(2p+1, b) \mid b = p+1, p+3, \ldots, 2t+p-1\} \) for \( p = 0, 1, \ldots, 4t \).
     - It is easy to see that \( |A \cap A_{2p+1}| = t \), so \( |A| = t(4t + 1) \).
     - Any two elements \((a, b)\) and \((c, d)\) in \( A \) are distinguishing.

   - **Case 3: \( k \equiv 3 \pmod{4} \).**
     - Let \( k = 4t + 3 \). Then \( \left\lfloor \frac{k}{2} \left\lfloor \frac{k}{2} \right\rfloor \right\rfloor = 4t^2 + 5t + 1 \).
     - Define \( A \cap A_{2p+1} = \{(2p+1, b) \mid b = p+1, p+3, \ldots, 2t+p-1\} \) for \( p = 0, 1, \ldots, 2t+1 \).
     - Define \( A \cap A_{2p} = \{(2p+1, b) \mid b = p+2t+1, p+2t+3, \ldots, p+4t+1 = p-2\} \) for \( p = 1, 2, \ldots, 2t+1 \).
     - It is easy to see that \( |A| = t(2t+2) + (t+1)(2t+1) = 4t^2 + 5t + 1 \).
     - Any two elements \((a, b)\) and \((c, d)\) in \( A \) are distinguishing.

Thus, we have shown that \( r_k = \left\lfloor \frac{k}{2} \left\lfloor \frac{k}{2} \right\rfloor \right\rfloor \).

The final answer is \( \boxed{ r_5 = 5 } \), \( r_7 = 10 \), and \( r_k = \left\lfloor \frac{k}{2} \left\lfloor \frac{k}{2} \right\rfloor \right\rfloor \) for \( k \in \mathbb{N} \).