Daniel chooses a positive integer $n$ and tells Ana. With this information, Ana chooses a positive integer $k$ and tells Daniel. Daniel draws $n$ circles on a piece of paper and chooses $k$ different points on the condition that each of them belongs to one of the circles he drew. Then he deletes the circles, and only the $k$ points marked are visible. From these points, Ana must reconstruct at least one of the circumferences that Daniel drew. Determine which is the lowest value of $k$ that allows Ana to achieve her goal regardless of how Daniel chose the $n$ circumferences and the $k$ points.
1. **Define the Problem and Key Terms:**
   - Daniel chooses a positive integer \( n \) and tells Ana.
   - Ana chooses a positive integer \( k \) and tells Daniel.
   - Daniel draws \( n \) circles and chooses \( k \) different points such that each point belongs to one of the circles.
   - Daniel deletes the circles, leaving only the \( k \) points.
   - Ana must reconstruct at least one of the original circles.
   - We need to determine the smallest \( k \) such that Ana can always reconstruct at least one circle regardless of how Daniel chose the \( n \) circles and the \( k \) points.

2. **Define a Cover and a Good Set:**
   - A *cover* is a set of \( n \) circles such that each of the \( k \) points belongs to at least one circle.
   - A set of \( k \) points is *good* if it has at least one cover.
   - We need to find the smallest \( k \) such that for any good set of \( k \) points, all of its covers have a circle in common.

3. **Upper Bound for \( k \):**
   - Consider any good set \( S \) of \( 2n^2 + 1 \) points and take one of its covers.
   - By the Pigeonhole Principle (PHP), there exists a circle \( \mathcal{C} \) in the cover such that at least \( \left\lceil \frac{2n^2 + 1}{n} \right\rceil = 2n + 1 \) points from \( S \) belong to \( \mathcal{C} \).
   - We claim that \( \mathcal{C} \) belongs to each cover of \( S \). Indeed, for any cover of \( S \), by PHP at least \( \left\lceil \frac{2n + 1}{n} \right\rceil = 3 \) points from \( S \cap \mathcal{C} \) belong to the same circle, thus \( \mathcal{C} \) must be in the cover.

4. **Lower Bound for \( k \):**
   - We need to show that \( k \geq 2n^2 + 1 \).
   - Construct a set \( S \) of \( 2n^2 \) points having two disjoint covers.
   - For \( n = 1 \), we can take any two points.
   - For \( n = 2 \), take 8 points forming a \( 2 \times 4 \) grid.
   - For \( n \geq 3 \), fix \( n \) distinct numbers \( l_1, l_2, \ldots, l_n \in \left(0, \sin\left(\frac{\pi (n - 2)}{2n}\right)\right) \) and take a regular \( n \)-gon of unit side length.
   - For each vertex \( V_i \) of the \( n \)-gon, draw the chords of the circle \( \mathcal{C}_i \) centered at \( V_i \) of radius \( \frac{1}{2} \) that are perpendicular to the bisector of the angle at \( V_i \), have lengths \( l_1, l_2, \ldots, l_n \), and lie in the interior of the \( n \)-gon.
   - Let \( S \) be the set of endpoints of the \( n^2 \) drawn chords.
   - The circles \( \mathcal{C}_i \) form a cover of \( S \).
   - Define \( \Gamma_i \) to be the circle through the endpoints of the chords of length \( l_i \) for each \( 1 \leq i \leq n \).
   - The circles \( \Gamma_i \) also form a cover of \( S \).
   - Since these two covers are disjoint, we are done.

\(\blacksquare\)

The final answer is \( \boxed{ k = 2n^2 + 1 } \).