Let $n$ be a given positive integer. In the Cartesian plane, each lattice point with nonnegative coordinates initially contains a butterfly, and there are no other butterflies. The [i]neighborhood[/i] of a lattice point $c$ consists of all lattice points within the axis-aligned $(2n+1) \times (2n+1)$ square entered at $c$, apart from $c$ itself. We call a butterfly [i]lonely[/i], [i]crowded[/i], or [i]comfortable[/i], depending on whether the number of butterflies in its neighborhood $N$ is respectively less than, greater than, or equal to half of the number of lattice points in $N$. Every minute, all lonely butterflies fly away simultaneously. This process goes on for as long as there are any lonely butterflies. Assuming that the process eventually stops, determine the number of comfortable butterflies at the final state.
1. **Understanding the Problem:**
   - Each lattice point with nonnegative coordinates initially contains a butterfly.
   - The neighborhood of a lattice point \( c \) consists of all lattice points within the axis-aligned \((2n+1) \times (2n+1)\) square centered at \( c \), excluding \( c \) itself.
   - A butterfly is lonely if the number of butterflies in its neighborhood is less than half of the number of lattice points in the neighborhood.
   - The process continues until there are no lonely butterflies.

2. **Characterizing the Final State:**
   - The final state is a regular set of butterflies, none of which are lonely.
   - The optimal state is defined as the state consisting of all points on or above the optimal curve, which is derived from the Farey sequence of order \( n+1 \).

3. **Proving the Optimal State is a Final State:**
   - **Vanguards on the Convex Hull are Comfortable:**
     - For any vanguard \( A \) on the convex hull, every lattice point to the left and below \( A \) is unoccupied, and every point to the right and above is occupied.
     - The number of lattice points in the neighborhood of \( A \) is equal to the number of occupied points, making \( A \) comfortable.
   - **Vanguards not on the Convex Hull are Crowded:**
     - For a vanguard \( V \) not on the convex hull, consider the nearest vanguards \( A \) and \( B \) on the convex hull.
     - \( V \) has at least as many neighbors as \( A \) and \( B \), making \( V \) crowded.

4. **Every Final State is a Subset of the Optimal State:**
   - Consider the convex hull of an arbitrary final state.
   - The sequence of gradients of the segments of the convex hull must be strictly increasing.
   - Each section of the broken line formed by the vanguards on the convex hull must lie within a lattice square of side length \( n+1 \).
   - The convex hull of the final state is a subset of the optimal state.

5. **If a Final State is a Proper Subset of Another, the Process Cannot End in the Subset:**
   - For any final state \( X \), every butterfly in \( X \) is present at the start.
   - If all butterflies in \( X \) are present at some time \( t \), they will be present at \( t+1 \) and hence at the termination of the process.
   - Therefore, the process must end in the optimal state.

6. **Counting the Comfortable Butterflies in the Optimal State:**
   - Overlaying the segments of the optimal curve such that their upper-left endpoints are at the origin, every lattice point in the square with corners \((-1,-1)\) and \((-(n+1),-(n+1))\) lies on exactly one segment.
   - Each of these points corresponds to exactly one convex hull vanguard of the optimal state.
   - The number of comfortable butterflies is \( n^2 + 1 \).

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