Let the vertices be $P_1,...,P_{2n+1}$ with indices $\mod (2n+1)$. For $k=1,...,2n+1$, we'll label $P_k$ by $1+2kn\mod {(4n+2)}$ and the midpoint of $P_kP_{k+1}$ by $2k$. I'll leave to the reader to check that this is well-defined, the vertices get each odd number exactly once and the sum for each side is $6n+4$.

Let us denote the vertices of the polygon as $V_1, V_2, .. V_{2n+1}$ in clockwise manner. Start assigning consecutive positive numbers $1, 2, 3 ... 2n+1$ to every second vertex starting with vertex $V_1$ in clockwise manner. The idea being that, with this arrangement sum of $2$ numbers on each side increases by $1$. (Sums are consecutive numbers in increasing order). Now, all that remains is to add remaining consecutive positive numbers in decreasing order to the above sequence so that the differences in sums between each side becomes $0$. So start assigning consecutive positive numbers $4n+2, 4n+1, 4n, ... 2n+2$ in clockwise manner to mid-points of each side starting with midpoint of side $V_{2n+1}V_1$. This results in sum of all $3$ numbers of each side of the polygon being equal to $5n+4$.