The answer is $n \ge 3$. We abbreviate PP = polygonal pythagorean.

First, assume that $n$ is PP. We will show then that $n + 2$ is also PP. Let $a_1, \dots, a_n$ be pairwise different positive integers such that $a_i^2 + a_{i+1}^2$ is a perfect square for all $i = 1, \dots, n$, where $a_{n+1} = a_1$. Choose any Pythagorean triple $(x, y, z)$, that is, three positive integers such that $x^2 + y^2 = z^2$. We claim that $a_1x, a_2x, \dots, a_nx, a_ny, a_1y$ satisfy the desired conditions for $n + 2$. Indeed,
$$\begin{align*} (a_i x)^2 + (a_{i+1} x)^2 &= x^2 (a_i^2 + a_{i+1}^2), \\ (a_n x)^2 + (a_n y)^2 &= a_n^2 (x^2 + y^2), \\ (a_n y)^2 + (a_1 y)^2 &= y^2 (a_n^2 + a_1^2), \quad \text{and} \\ (a_1 y)^2 + (a_1 x)^2 &= a_1^2 (y^2 + x^2), \end{align*}$$
which are all products of two perfect squares and therefore are perfect squares themselves.

However, it may happen that either $a_ny$ or $a_1y$ is equal to some $a_ix$. To make sure that this is not the case, take a prime number $p$ which does not divide any of the $a_i$'s, and use the Pythagorean triple $(x, y, z) = (p^2 - 1, 2p, p^2 + 1)$. Since $y$ is divisible by $p$ and both $a_i$ and $x$ are not, no $a_ix$ can be equal to $a_ny$ or $a_1y$, and so the $n+2$ numbers $a_1x, a_2x, \dots, a_nx, a_ny, a_1y$ are pairwise different, which proves our claim.

With the example given in the problem statement we are able to get solutions for all odd $n$. To solve the problem for even $n$, it is enough to find a solution for $n = 4$. Considering the Pythagorean triples $(3, 4, 5)$ and $(5, 12, 13)$, we can check that $(3 \cdot 5, 4 \cdot 5, 4 \cdot 12, 3 \cdot 12) = (15, 20, 48, 36)$ is a solution, and the proof is complete.