We will solve the following stronger problem. Problem. Show that there is infinitely many distinct hexagons in the plane such that the distance between every pair of vertices is an integer. (We call two hexagons in the plane distinct if one cannot be transformed into the other by rotation, translation, scaling or reflection.) Solution The following result appeared on the 2020 New Zealand TST. Lemma. The equation $$ c^2 = a^2 + ab + b^2 $$has infinitely many positive integer solutions such that $a$, $b$ and $c$ have no common factor greater than 1. Proof. For $n \geq 2$ let \[ a=2 n+1, \quad b=n^{2}-1, \quad c=n^{2}+n+1 \]Then \[ \begin{aligned} a^{2}+ ab +b^{2} &=(2 n+1)^{2}+\left(n^{2}-1\right)^{2}+(2 n+1)\left(n^{2}-1\right) \\ &=\left(4 n^{2}-4 n+1\right)+\left(n^{4}-2 n^{2}+1\right)+\left(2 n^{3}+n^{2}-2 n-1\right) \\ &=n^{4}+2 n^{3}+3 n^{2}+2 n+1 \\ &=\left(n^{2}+n+1\right)^{2} \\ &=c^{2} \end{aligned} \]so $a$, $b$, $c$ are a positive integer solution to the equation. To show that there are infinitely many solutions with no common factor greater than 1, observe that any prime factor of $a, b$ and $c$ would also have to divide \[ 2 c-2 a-b=3 \]We can avoid a common factor of 3 by choosing $n \neq 1 \pmod{3}$. Since there are infinitely many such $n$ this proves that there are infinitely many relatively prime solutions, as required. Now with this proven, we can go back to our main problem. Consider the equiangular hexagon $ABCDEF$ with alternating side lengths $a$ and $b$, and diagonals of length $c$ and $d$ as shown below. To solve our problem, it suffices to show that there are infinitely many pairs of positive integers $a$, $b$, $c$ and $d$ such that $\gcd(a, b, c, d) = 1$, that can be formed into a hexagon such as the one shown above. Considering the triangle $ABC$, we can apply the cosine rule to get $$ c^2 = a^2 + b^2 - 2ab \cos 120^\circ \implies c^2 = a^2 + ab + b^2, $$so $a$, $b$ and $c$ satisfy this equation. The quadrilateral $ABCF$ has $\angle FAB + \angle BCF = 120^\circ + 60^\circ = 180^\circ$, so it must be cyclic. Thus we can apply Ptolemy's theorem to get $$ a^2 + bd = c^2 \implies d = \frac{c^2 - a^2}{b}. $$If we want $d$ to be an integer, then we must have $b \mid c^2 - a^2$, but $c^2 - a^2 = ab + b^2$, so, if $a$, $b$ and $c$ are positive integers, then $b \mid c^2 - a^2$ so $d$ is also an integer. So, given any positive integer solution to $c^2 = a^2 + ab + b^2$ we can construct a hexagon where the distance between every pair of vertices is an integer. However, with out lemma, we proved that this equation has infinitely many solutions with no common factor greater than 1, hence we can create infinitely many distinct hexagons with this property, finishing our solution.

The solution above also works to generate suitable hexagons, by using the parameterisation given in the proof of the lemma. Some examples are shown below.

An alternate solution (not mine) involving pythagorean triples. Solution Consider the following hexagon, with $x$, $y$ and $z$ all positive integers. To have an integer distance between every pair of vertices, we need the following to all be perfect squares. $$ x^2 + y^2, \quad \quad y^2 + (x + z)^2, \quad \text{and} \quad z^2 + 4y^2. $$Checking by hand, we find that $x = 9$, $y = 12$ and $z = 7$ works, which gives the hexagon below.

My solution is not as nice as above. Please correct me if I am wrong. Just a sketch

This problem (or at least an equivalent problem) also appeared in the well known collection of 'jewish problems', as problem 12 (https://arxiv.org/pdf/1110.1556.pdf). The paper contains the following hint and solution. Hint: A Pythagorean triangle generates three such points. Various reflections of it can increase the number of points. Solution: If we take a right triangle $A B C$ with integer sides, the area will be rational. Therefore, the height onto the hypotenuse $A C$ will be rational. This means that if we reflect this triangle about the hypotenuse, we will have four points in the plane all at rational distances from each other. Now, let the base of that height be at point $D$. Then, since the triangle $A B C$ is similar to the triangle $B D C,$ the distances $A D$ and $C D$ will both be rational. Therefore, reflecting the whole thing about the perpendicular bisector of $A C$ yields two more points, so that all six are at rational distances from each other. All that's left now is to expand by a multiplicative factor that clears all the denominators. Start with the 3,4,5 Pythagorean triangle, lay the hypotenuse along the $x$ axis, with the center at the origin, perform the above construction, and scale by 10 to clear denominators. Six such points are: \[ (\pm 25,0),\quad (\pm 7,\pm 24). \]

Is there an n-gon for every $n \ge 3$ with integer distances between all vertices?