Let $ n \geq 3$ and consider a set $ E$ of $ 2n \minus{} 1$ distinct points on a circle. Suppose that exactly $ k$ of these points are to be colored black.  Such a coloring is [b]good[/b] if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly $ n$ points from $ E$.  Find the smallest value of $ k$ so that every such coloring of $ k$ points of $ E$ is good.
To solve this problem, we need to determine the smallest value of \( k \) such that any coloring of \( k \) points out of \( 2n-1 \) points on a circle is "good." A coloring is "good" if there is at least one pair of black points such that the interior of one of the arcs between them contains exactly \( n \) points from \( E \).

1. **Graph Representation**:
   - Consider the \( 2n-1 \) points on the circle as vertices of a graph.
   - Connect two vertices if there are exactly \( n \) vertices between them on the circle (in either of the two arcs).

2. **Degree of Each Vertex**:
   - Each vertex will have a degree of 2 because each vertex will be connected to exactly two other vertices (one in each direction around the circle).

3. **Structure of the Graph**:
   - The graph is composed of disjoint cycles. Each cycle is formed by vertices that are \( n \) points apart.

4. **Case Analysis**:
   - **Case 1: \( 2n-1 \) is not divisible by 3**:
     - In this case, the graph consists of a single cycle of length \( 2n-1 \).
     - The maximum size of an independent set in a cycle of length \( 2n-1 \) is \( \left\lfloor \frac{2n-1}{2} \right\rfloor = n-1 \).
     - Therefore, if we color \( n \) vertices, there must be at least one pair of consecutive black vertices, ensuring that the coloring is "good."

   - **Case 2: \( 2n-1 \) is divisible by 3**:
     - In this case, the graph is the union of 3 disjoint cycles, each of length \( \frac{2n-1}{3} \).
     - The maximum size of an independent set in each cycle of length \( \frac{2n-1}{3} \) is \( \left\lfloor \frac{\frac{2n-1}{3}}{2} \right\rfloor = \frac{2n-1}{6} \).
     - Therefore, the total maximum size of an independent set in the entire graph is \( 3 \times \frac{2n-1}{6} = \frac{2n-1}{2} = n-1 \).
     - Thus, if we color \( n \) vertices, there must be at least one pair of consecutive black vertices in at least one of the cycles, ensuring that the coloring is "good."

5. **Conclusion**:
   - In both cases, the smallest value of \( k \) such that every coloring of \( k \) points is "good" is \( k = n \).

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