Let $k$ be a positive integer and $S$ be a set of sets which have $k$ elements. For every $A,B \in S$ and $A\neq B$ we have $A \Delta B \in S$. Find all values of $k$ when $|S|=1023$ and $|S|=2023$.

Note:$A \Delta B = (A \setminus B) \cup (B \setminus A)$
1. **Understanding the Symmetric Difference**:
   Given \( A \Delta B = (A \setminus B) \cup (B \setminus A) \), we can rewrite it as \( A \oplus B \). Notice that if \( C = A \oplus B \), then \( B = A \oplus C \) and \( A = C \oplus B \).

2. **Cardinality and Symmetric Difference**:
   Since \( |A| = |B| = |A \oplus B| = k \), it implies that \( |A \cap B| = \frac{k}{2} \). Therefore, \( k \) must be even for \( |S| \geq 2 \).

3. **Odd Cardinality of \( S \)**:
   Let \( A \in S \). For any other element \( X \) in \( S \), there will be another unique set \( X' \in S \) such that \( A \oplus X = X' \), \( A \oplus X' = X \), and \( X \oplus X' = A \). This implies that we can pair up the elements of \( S \setminus \{A\} \) in pairs of \( (X, X') \). Hence, \( |S| \) must be odd.

4. **Form of \( |S| \)**:
   Let \( |S| = 2n + 1 \) and \( k = 2m \). We can construct pairs \( (X, X') \) such that \( a \not\in X \) and \( a \in X' \). Let the set of pairs be \( T = \{(X_1, X_1'), \dots, (X_n, X_n')\} \).

5. **Claim 1**:
   The set \( S' = \{X_i \mid (X_i, X_i') \in T\} \) is closed under \( \oplus \).
   - **Proof**: Here, \( X_i \oplus X_j \in S \) and \( a \not \in X_i \oplus X_j \implies X_i \oplus X_j \not\in \{A, X_1', X_2', \dots, X_n'\} \). So, \( X_i \oplus X_j \in S' \). \(\square\)

6. **Claim 2**:
   The set \( S'' = \{X_i \cap A \mid X_i \in S'\} \) is closed under \( \oplus \).
   - **Proof**: It's trivial that \( (X_i \cap A) \oplus (X_j \cap A) = (X_i \oplus X_j) \cap A \) and from Claim 1 we get, \( (X_i \oplus X_j) \in S' \implies (X_i \oplus X_j) \cap A \in S'' \). \(\square\)

7. **Implications**:
   If \( k \) works for \( |S| = 2n + 1 \), then \( k \) works for \( |S| = n \). Since \( |S| \) is always odd, \( |S| \) must be \( 2^d - 1 \) for some \( d \). By Claim 2, if \( k \) works for \( |S| = 2n + 1 \), then \( \frac{k}{2} \) works for \( |S| = n \) and \( 2 \mid k \) for \( |S| > 1 \). Therefore, if \( |S| = 2^d - 1 \), then \( 2^{d-1} \mid k \).

8. **Conclusion**:
   - For \( |S| = 1023 \), \( 1023 = 2^{10} - 1 \), so \( 2^9 \mid k \).
   - For \( |S| = 2023 \), there is no solution since \( 2023 \) is not of the form \( 2^d - 1 \).

The final answer is \( \boxed{ k = 2^9 \times m } \) for \( |S| = 1023 \) and no solution for \( |S| = 2023 \).