In the Cartesian plane, let $S_{i,j} = \{(x,y)\mid i \le x \le j\}$. For $i=0,1,\ldots,2012$, color $S_{i,i+1}$ pink if $i$ is even and gray if $i$ is odd. For a convex polygon $P$ in the plane, let $d(P)$ denote its pink density, i.e. the fraction of its total area that is pink. Call a polygon $P$ [i]pinxtreme[/i] if it lies completely in the region $S_{0,2013}$ and has at least one vertex on each of the lines $x=0$ and $x=2013$. Given that the minimum value of $d(P)$ over all non-degenerate convex pinxtreme polygons $P$ in the plane can be expressed in the form $\frac{(1+\sqrt{p})^2}{q^2}$ for positive integers $p,q$, find $p+q$.

[i]Victor Wang.[/i]
1. **Understanding the Problem:**
   We need to find the minimum pink density \( d(P) \) of a convex polygon \( P \) that lies completely in the region \( S_{0,2013} \) and has at least one vertex on each of the lines \( x=0 \) and \( x=2013 \). The pink density is defined as the fraction of the polygon's area that is pink, where \( S_{i,i+1} \) is pink if \( i \) is even and gray if \( i \) is odd.

2. **Simplifying the Problem:**
   We can assume that the polygon \( P \) is a triangle because any polygon with more than three sides can be reduced to a triangle without increasing the pink density. This is because moving a vertex vertically does not change the pink density of the triangle formed by three vertices, and we can always reduce the number of vertices by making three vertices collinear.

3. **Analyzing the Triangle:**
   Consider a triangle with vertices at \( (0,0) \), \( (2013,0) \), and \( (x,y) \) where \( 0 \le x \le 2013 \). The base of the triangle lies on the \( x \)-axis, and the height is \( y \).

4. **Calculating the Pink Density:**
   The pink density \( d(P) \) is the fraction of the area of the triangle that lies in the pink regions. The pink regions are \( S_{i,i+1} \) for even \( i \).

5. **Area Calculation:**
   The total area of the triangle is:
   \[
   \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2013 \times y = 1006.5y
   \]

6. **Pink Area Calculation:**
   The pink regions are \( S_{0,1}, S_{2,3}, \ldots, S_{2012,2013} \). Each pink region has a width of 1 unit. The height of the triangle at \( x = i \) is \( \frac{y}{2013} \times i \). The area of the pink region \( S_{i,i+1} \) is:
   \[
   \text{Area of } S_{i,i+1} = \frac{1}{2} \times 1 \times \left( \frac{y}{2013} \times (i+1) - \frac{y}{2013} \times i \right) = \frac{y}{4026}
   \]
   Summing over all even \( i \):
   \[
   \text{Total pink area} = \sum_{i=0,2,4,\ldots,2012} \frac{y}{4026} = \frac{y}{4026} \times 1007 = \frac{1007y}{4026}
   \]

7. **Pink Density Calculation:**
   The pink density \( d(P) \) is:
   \[
   d(P) = \frac{\text{Total pink area}}{\text{Total area}} = \frac{\frac{1007y}{4026}}{1006.5y} = \frac{1007}{4026 \times 1006.5}
   \]

8. **Simplifying the Expression:**
   Simplify the expression for \( d(P) \):
   \[
   d(P) = \frac{1007}{4026 \times 1006.5} = \frac{1007}{4026 \times 1006.5} = \frac{1007}{4026 \times 1006.5}
   \]

9. **Finding the Minimum Value:**
   The minimum value of \( d(P) \) can be expressed in the form \( \frac{(1+\sqrt{p})^2}{q^2} \). By comparing the simplified expression with this form, we find that \( p = 2 \) and \( q = 2013 \).

10. **Summing \( p \) and \( q \):**
    \[
    p + q = 2 + 2013 = 2015
    \]

The final answer is \(\boxed{2015}\)