Throughout this solution, we remove the units (cm) as each length is in these same units. First, we calculate the distance flown by the fly, which we call $f$. Let $Z$ be the point on the base on the prism directly underneath $Y$. Since the hexagonal base has side length 30, then $XZ = 60$. This is because a hexagon is divided into 6 equilateral triangles by its diagonals, and so the length of the diagonal is twice the side length of one of these triangles, which is twice the side length of the hexagon. Also, $\triangle XZY$ is right-angled at $Z$, since $XZ$ lies in the horizontal base and $YZ$ is vertical. By the Pythagorean Theorem, since $XY > 0$, then $XY = \sqrt{XZ^{2} + YZ^{2}} = \sqrt{60^{2} + 165^{2}}$. Therefore, $f = XY = \sqrt{60^{2} + 165^{2}}$. Next, we calculate the distance crawled by the ant, which we call $a$. Since the ant crawls $n + \frac{1}{2}$ around the prism and its crawls along all 6 of the vertical faces each time around the prism, then it crawls along a total of $6(n + \frac{1}{2}) = 6n + 3$ faces. To find $a$, we 'unwrap' the exterior of the prism. Since the ant passes through $6n + 3$ faces, it travels a 'horizontal' distance of $(6n + 3) \cdot 30$. Since the ant moves from the bottom of the prism to the top of the prism, it passes through a vertical distance of 165. Since the ant's path has a constant slope, its path forms the hypotenuse of a right-angled triangle with base of length $(6n + 3) \cdot 30$ and height of length 165. By the Pythagorean Theorem, since $a > 0$, then $a = \sqrt{((6n + 3) \cdot 30)^{2} + 165^{2}}$. Now, we want $a$ to be at least $20f$. In other words, we want to find the smallest possible value of $n$ for which $a > 20f$. Since these quantities are positive, the inequality $a > 20f$ is equivalent to the inequality $a^{2} > 20^{2}f^{2}$. The following inequalities are equivalent: $(6n + 3)^{2} \cdot 2^{2} + 11^{2} > 400(4^{2} + 11^{2})$, $4(6n + 3)^{2} + 121 > 400 \cdot 137$, $4(6n + 3)^{2} > 54679$, $(6n + 3)^{2} > \frac{54679}{4}$, $6n + 3 > \sqrt{\frac{54679}{4}}$ (since both sides are positive), $6n > \sqrt{\frac{54679}{4}} - 3$, $n > \frac{1}{6}(\sqrt{\frac{54679}{4}} - 3) \approx 18.986$. Therefore, the smallest positive integer $n$ for which this is true is $n = 19$.