Claim: An $n$-gon has a memorable labelling of consecutive integers iff $n\equiv 0$ (mod 4). Let us find some necessary conditions on such labelings first... Note that a memorable labelling remains memorable if any integer constant $c$ is added to all numbers. Thus, w.l.o.g., from now on we only consider memorable labellings that use the $2n+1$ consecutive numbers $\{0,\ldots, 2n\}$. The minimum $0$ and the maximum $2n$ are surely used to label vertices of the $n$-gon (since the sides and the interior are labelled with arithmetic means of the vertex values). Furthermore, all vertex labels are the same parity since one even and one odd value at neighboring vertices would lead to a non-integer label of its connecting side. Thus, the $n$ vertices of the $n$-gon are labelled with exactly $n$ of the $n+1$ even numbers of set $\{0,2,4,\ldots, 2n-2,2n\}$. The sum of all $n$ vertex values needs to be divisibly by $n$ (since the interior value needs to be integer) The sum of all available $n+1$ even numbers $\Sigma_{i=0}^n 2i = n(n+1)$ is divisible by $n$. Thus the only even number that is not used as a vertex number Needs to be divisible by $n$. As $0$ and $2n$ are used as vertex numbers, the only even number that is no vertex number is $n$. Then the sum of all vertex numbers is $n(n+1)-n=n^2$ and its average is $n$. We found that it is a necessary condition for the interior number to equal $n$. Case 1: $n$ odd The interior number $n$ is odd, so the sum $n^2$ of the vertex numbers is odd as well. Contradiction to: all vertex numbers are even. Case 2: $n\equiv 2$ mod 4 The numbers ${0,2,\ldots,n-2,n+2,\ldots 2n}$ are used as vertex numbers, $n$ is the interior number, all side numbers are therefore necessarily odd. We have $n/2+1$ vertex numbers that are $\equiv 0$ (mod 4) (namely $0,4,\ldots,2n$) and only $n/2-1$ vertex numbers that are $\equiv 2$ (mod 4) (namely $2,4,\ldots, 2n-2$, excluding $n$). Thus two neighboring vertex numbers exist that are both $\equiv 0$ (mod 4) and the connecting side has an even value which is a contradiction. Case 3: $n\equiv 0$ mod 4 Label the interior of the $n$-gon with $n=4k$ and the vertices consecutively as follows: $0, 2, 4, 6, \ldots, 4k-6, 4k-4, 4k-2, 4k+4, 4k+2, 4k+8, 4k+6, \ldots, 8k-4, 8k-6, 8k, 8k-2$ (i.e., start with the arithmetic series till $4k-2$, then alternatively add $6$ and subtract $2$ till you reach $8k-2$, closing the circle) The sides are then labelled with $1,3,5,7,\ldots, 4k-5, 4k-3, 4k+1, 4k+3, 4k+5, \ldots, 8k-5, 8k-3, 8k-1, 4k-1$ (i.e., an arithmetic series, except for $4k-1$ which appears as average of the vertices labelled $8k-2$ and $0$).

euklid wrote: Thus, the $n$ vertices of the $n$-gon are labelled with exactly $n$ of the $n+1$ even numbers of set $\{0,2,4,\ldots, 2n-2,2n\}$. Why the vertices can't be (all)odd?

reveryu wrote: euklid wrote: Thus, the $n$ vertices of the $n$-gon are labelled with exactly $n$ of the $n+1$ even numbers of set $\{0,2,4,\ldots, 2n-2,2n\}$. Why the vertices can't be (all)odd? Because the minimum ($0$) and maximum ($2n$) have to be used among the vertices.

Trivial stuff first. We can shift the numbers to $0, 1, 2, \cdots 2n$. The maximum number $2n$ and the minimum number $0$ must lie on the vertex. All numbers on the vertex have equal parity, being even numbers. The sum of all numbers is $2n^2+n$, so the center is $n$, and the sum of all vertex/side is $n^2$. First, we show that $n$ cannot be odd. If $n$ is odd, the sum of all vertex is odd, but the number of each vertex is even. We show that $n$ cannot be $2 \pmod{4}$. If $n=4k+2$, we have $2k+2$ numbers which is a multiple of $4$, and $2k+1$ numbers which is $2 \pmod{4}$. Now on the vertices, we have $2k+2$ numbers which is a multiple of $4$, and $2k$ numbers which is $2 \pmod{4}$. This is because the center has a number $2 \pmod{4}$. Since all numbers on the side must be odd, no consecutive numbers on the vertex can be same mod 4. This is impossible. Now we show that all $n$ which is a multiple of $4$ works. My construction is the same as #2.