Solution:
Answer: $n = 9$

For $n < 6$, there is at most one length of diagonal. For $n = 6$, $7$ the longest and shortest, and a side of the $n$-gon form a triangle, so the difference between the longest and shortest is less than the side.

For $n > 7$ the side has length $2R \sin \dfrac{\pi}{n}$, the shortest diagonal has length $2R \sin \dfrac{2\pi}{n}$, and the longest diagonal has length $2R$ for $n$ even and $2R \cos \dfrac{\pi}{2n}$ for $n$ odd (where $R$ is the radius of the circumcircle). Thus we require:

$\sin \dfrac{2\pi}{n} + \sin \dfrac{\pi}{n} = 1$ and $n$ even, or

$\sin \dfrac{2\pi}{n} + \sin \dfrac{\pi}{n} = \cos \dfrac{\pi}{2n}$ and $n$ odd.

Evidently the lhs is a strictly decreasing function of $n$ and the rhs is an increasing function of $n$, so there can be at most one solution of each equation. The second equation is satisfied by $n = 9$, although it is easier to see that there is a quadrilateral with the longest diagonal and shortest diagonals as one pair of opposite sides, and $9$-gon sides as the other pair of opposite sides. The angle between the longest side and an adjacent side is $60$, so that its length is the length of the shortest diagonal plus $2 \times$ $9$-gon side $\times \cos 60$. Hence that is the only solution for $n$ odd.

For $n = 8$ we have the same quadrilateral as for the $9$-gon except that the angle is $67.5$ and hence the difference is less than $1$. For $n = 10$, $\sin \dfrac{2\pi}{10} + \sin \dfrac{\pi}{10} = \sin \dfrac{\pi}{10}(2 \cos \dfrac{\pi}{10} + 1) < 3 \sin \dfrac{\pi}{10} < 3\dfrac{\pi}{10} < 1$. So there are no solutions for $n$ even $\geq 10$, and hence no solutions for $n$ even.