Let $n$ be a positive integer. Find the smallest positive integer $k$ such that for any set $S$ of $n$ points in the interior of the unit square, there exists a set of $k$ rectangles such that the following hold:
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[*]The sides of each rectangle are parallel to the sides of the unit square.
[*]Each point in $S$ is [i]not[/i] in the interior of any rectangle.
[*]Each point in the interior of the unit square but [i]not[/i] in $S$ is in the interior of at least one of the $k$ rectangles
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(The interior of a polygon does not contain its boundary.)

[i]Holden Mui[/i]
To solve this problem, we need to find the smallest positive integer \( k \) such that for any set \( S \) of \( n \) points in the interior of the unit square, there exists a set of \( k \) rectangles satisfying the given conditions. Let's break down the problem step by step.

1. **Understanding the Problem:**
   - We have \( n \) points inside a unit square.
   - We need to cover the entire interior of the unit square with \( k \) rectangles such that:
     - The sides of each rectangle are parallel to the sides of the unit square.
     - Each point in \( S \) is not in the interior of any rectangle.
     - Each point in the interior of the unit square but not in \( S \) is in the interior of at least one of the \( k \) rectangles.

2. **Initial Observations:**
   - If there is only one point in \( S \), we can cover the unit square with 2 rectangles by splitting the square vertically or horizontally, avoiding the point.
   - For \( n \) points, we need to generalize this idea.

3. **General Strategy:**
   - For each point in \( S \), we can create two rectangles to cover the space around the point while avoiding the point itself.
   - This suggests that for \( n \) points, we might need \( 2n \) rectangles.

4. **Detailed Calculation:**
   - Consider a point \( P_i \) in \( S \) with coordinates \((x_i, y_i)\).
   - We can create two rectangles:
     - One covering the area to the left of \( P_i \) and extending vertically from the bottom to the top of the square.
     - Another covering the area to the right of \( P_i \) and extending vertically from the bottom to the top of the square.
   - This ensures that \( P_i \) is not in the interior of either rectangle.

5. **Covering the Entire Square:**
   - To ensure that every point in the interior of the unit square but not in \( S \) is covered, we need to extend this idea to all \( n \) points.
   - For each point, we need two rectangles, one on each side of the point.

6. **Bounding the Number of Rectangles:**
   - For \( n \) points, we need \( 2n \) rectangles to cover the space around each point.
   - However, we need to ensure that the entire interior of the unit square is covered without any gaps.

7. **Final Adjustment:**
   - To cover the entire interior of the unit square, we need to add two more rectangles to cover the remaining uncovered areas at the boundaries.
   - Therefore, the total number of rectangles required is \( 2n + 2 \).

Conclusion:
The smallest positive integer \( k \) such that for any set \( S \) of \( n \) points in the interior of the unit square, there exists a set of \( k \) rectangles satisfying the given conditions is \( \boxed{2n + 2} \).