Find the number of ordered pairs $(x,y)$ of positive integers that satisfy
$$ x \leq 2y \leq 60, \quad y \leq 2x \leq 60. $$
The solution set of these inequalities is the quadrilateral with vertices at $(0,0), (15,30), (30,30), (30,15)$. Since this quadrilateral has vertices at integer points, we use Pick's theorem, which states that $A = n - b/2 - 1$, where $A$ is the area of the quadrilateral, $n$ is the number of integer points in it (both boundary and interior), and $b$ is the number of integer points on the boundary. The quadrilateral can be constructed as a $30 \times 30$ square (area $900$) with two triangles removed that each have an area of $15 \cdot 30/2=225$, so the quadrilateral has area $450$. In addition to the vertices, the slanted sides that meet at $(0,0)$ have integer points at $(k,2k)$ and $(2k,k)$ for $k$ from $1$ to $14$, and the short sides that meet at $(30,30)$ have integer points at $(j,30)$ and $(30,j)$ for $j$ from $16$ to $29$, evaluating to a total of $60$ integer points on the boundary. Therefore we have $n = A + b/2 + 1 = 450 + 30 + 1 = 481$, and since we are only considering positive integer solutions we remove $(0,0)$ which leaves us with $480$ ordered positive integer pairs.
Huge thanks to Joan Llobera (Lleida, Catalonia), Mark Spindler (Maryland), Ibrahim (California), Kurt Wynn (Brockton Bay), David J. Webb (Hawaii), Darko L (Serbia), Martin Masson (France), Bethany (Texas), jc (New York), and Aayush Rath (India) for submitting solutions to this challenge problem.