Let $n,k$ be given natural numbers. Find the smallest possible cardinality of a set $A$ with the following property: There exist subsets $A_1,A_2,\ldots,A_n$ of $A$ such that the union of any $k$ of them is $A$, but the union of any $k-1$ of them is never $A$.
1. **Define the problem and notation:**
   We are given natural numbers \( n \) and \( k \). We need to find the smallest possible cardinality of a set \( A \) such that there exist subsets \( A_1, A_2, \ldots, A_n \) of \( A \) with the following properties:
   - The union of any \( k \) of these subsets is \( A \).
   - The union of any \( k-1 \) of these subsets is never \( A \).

2. **Claim:**
   We claim that the smallest possible cardinality of \( A \) is \( \binom{n}{n-k+1} \).

3. **Reasoning:**
   - For every element \( E \) in \( A \), \( E \) must appear in at least \( n-k+1 \) subsets \( A_i \) to ensure that it is included in the union of any \( k \) subsets.
   - \( E \) must also appear in at most \( n-k+1 \) subsets to ensure that it is not always included in the union of any \( k-1 \) subsets.

4. **Verification using the Pigeonhole Principle:**
   - If the elements of \( A \) are distributed among the subsets such that no two elements always appear together, the Pigeonhole Principle implies that at least one element does not appear in the union of every \( k-1 \) subset.
   - This condition is satisfied if and only if \( A \) has at least \( \binom{n}{n-k+1} \) elements.

5. **Optimality:**
   - Arranging the elements of \( A \) such that no two elements always appear together is optimal. If two elements always appear together, it effectively reduces the cardinality of \( A \) because those two elements behave identically with respect to the subsets.

6. **Conclusion:**
   - Therefore, the smallest possible cardinality of \( A \) is \( \binom{n}{n-k+1} \).

The final answer is \( \boxed{\binom{n}{n-k+1}} \).