A sequence $ (S_n), n \geq 1$ of sets of natural numbers with $ S_1 = \{1\}, S_2 = \{2\}$ and
\[{ S_{n + 1} = \{k \in }\mathbb{N}|k - 1 \in S_n \text{ XOR } k \in S_{n - 1}\}.
\]
Determine $ S_{1024}.$
To determine \( S_{1024} \), we need to understand the recursive definition of the sequence \( (S_n) \). The sequence is defined as follows:
\[ S_1 = \{1\}, \]
\[ S_2 = \{2\}, \]
\[ S_{n+1} = \{ k \in \mathbb{N} \mid (k-1 \in S_n) \text{ XOR } (k \in S_{n-1}) \}. \]

We will use induction to find a pattern in the sets \( S_n \).

1. **Base Cases:**
   \[ S_1 = \{1\}, \]
   \[ S_2 = \{2\}. \]

2. **Inductive Step:**
   Assume we have \( S_n \) and \( S_{n-1} \). We need to find \( S_{n+1} \).

   The definition of \( S_{n+1} \) is:
   \[ S_{n+1} = \{ k \in \mathbb{N} \mid (k-1 \in S_n) \text{ XOR } (k \in S_{n-1}) \}. \]

   This means \( k \in S_{n+1} \) if and only if exactly one of the following is true:
   - \( k-1 \in S_n \)
   - \( k \in S_{n-1} \)

3. **Pattern Recognition:**
   Let's compute the first few sets to identify a pattern:
   \[ S_3 = \{ k \in \mathbb{N} \mid (k-1 \in S_2) \text{ XOR } (k \in S_1) \}. \]
   Since \( S_2 = \{2\} \) and \( S_1 = \{1\} \):
   \[ S_3 = \{ k \in \mathbb{N} \mid (k-1 = 2) \text{ XOR } (k = 1) \}. \]
   \[ S_3 = \{3\}. \]

   \[ S_4 = \{ k \in \mathbb{N} \mid (k-1 \in S_3) \text{ XOR } (k \in S_2) \}. \]
   Since \( S_3 = \{3\} \) and \( S_2 = \{2\} \):
   \[ S_4 = \{ k \in \mathbb{N} \mid (k-1 = 3) \text{ XOR } (k = 2) \}. \]
   \[ S_4 = \{4\}. \]

   \[ S_5 = \{ k \in \mathbb{N} \mid (k-1 \in S_4) \text{ XOR } (k \in S_3) \}. \]
   Since \( S_4 = \{4\} \) and \( S_3 = \{3\} \):
   \[ S_5 = \{ k \in \mathbb{N} \mid (k-1 = 4) \text{ XOR } (k = 3) \}. \]
   \[ S_5 = \{5\}. \]

   We observe that \( S_n = \{n\} \) for \( n = 1, 2, 3, 4, 5 \).

4. **Generalization:**
   By induction, we can generalize that:
   \[ S_n = \{n\} \text{ for all } n \in \mathbb{N}. \]

5. **Conclusion:**
   Therefore, \( S_{1024} = \{1024\} \).

The final answer is \( \boxed{1024} \).