Let us assume we managed to place numbers properly. Let $m = \min_{j=1, n} a_j$, $M = \max_{j=1, n} a_j$. If $m = a_i = \frac{1}{2}(a_k + a_l)$, then $m = a_k = a_l$, hence there have to be at least 3 of the smallest numbers, similarly for the biggest. It is easy to fill in numbers when

Fig. 40

$n \ge 6$ is composite. Let $n = pq$, $p, q > 1$. Then we place $m$ in vertices $A_p, A_{2p}, \dots, A_{qp}$, and we place $M$ in the rest of the vertices. It is clear that we satisfy conditions since we split vertices of our polygon into $q$ regular $p$-sided polygons, with the same number at each vertex, and a number in each vertex of a polygon is the arithmetic mean of numbers in the rest of the vertices of the same polygon, hence we satisfy the condition for $n$-sided polygon.

Now we consider the case when $n$ is prime. Let us represent this number as $n = 4s + r$, where $r \in \{3, 5\}$. Next we propose the following arrangement of numbers in vertices of $n$-sided polygon.

First for $n=4s+3$. For the sake of simplification, on Fig. 40 we denote vertex $A_i$ as $i$, $m$ as black disk, and $M$ as a white disk. Now for every disk $i$ we have to indicate a pair of disks of the same color that are located symmetrically (on equal distance from $i$, hence with index $i+u$ and $i-u$ (mod $n$)). In case there are 3 or more consecutive disks of the same color, then we have to consider only those on the sides of the group since for the rest of them the desired pair is two adjacent disks. Therefore for the group of black vertices we have to find a pair for the disk with index $s+3$ (due to symmetry we have a similar answer for $3s+2$). It is obvious that a pair for $s+3$ will be 1 and a symmetric disk. We can calculate that its index is $2s+5$, since it has to satisfy $s+3 = \frac{1}{2}(1+2s+5)$. Hence it will be black if $2s+5 \le 3s+2$ or $s \ge 3$. Due to symmetry, for the white group it is enough to consider indices $2, 3, \dots, s+3$.

For 2 a pair is $4s+3$ and 4. Now we consider $s+2$. One could be a disk with the least possible index of $3s+3$. The other one has to have an index of
$$(s+2)-(2s+1)=1-s=1-s+4s+3=3s+4.$$
A distinct black vertex with index of 1 clearly satisfies conditions, for instance, its pair of symmetric vertices could be $2s+2$ and $2s+3$. Therefore this construction will do when $s \ge 3$, meaning $n=4s+3 \ge 15$.

Now we consider $n=4s+5$ (Fig. 41). For the group of black vertices we have to find a pair for the disk with index $s+4$. The desired pair is 1 and a symmetric disk with index of $2s+7$, since it has to satisfy $s+4 = \frac{1}{2}(1+2s+7)$. Hence it will be black when $2s+7 \le 3s+3$ or $s \ge 4$.

For the white group for 2 the pair is $4s+5$ and 4. Now we consider $s+3$. One could be a disk with the least possible index of $3s+4$. The other one has to have an index of
$$(s+3)-(2s+1)=2-s=2-s+4s+5=3s+7.$$
A distinct black vertex with index of 1 clearly satisfies conditions, its pair of symmetric vertices could be $2s+3$ and $2s+4$. Therefore this construction will do when $s \ge 3$, meaning $n=4s+5 \ge 21$.

Now we have to consider the following prime numbers: 17, 13, 11 and 7.
Let's present construction for the first three of them.

Have a look at Fig. 42 for $n=17$. Non-obvious pairs are for the following vertices:
$$\begin{aligned} 13 &= \frac{1}{2}(5+21) \equiv \frac{1}{2}(5+4), \\ 5 &= \frac{1}{2}(13-3) \equiv \frac{1}{2}(13+14), \\ 12 &= \frac{1}{2}(18+6) \equiv \frac{1}{2}(1+6), \\ 6 &= \frac{1}{2}(1+11). \end{aligned}$$

Have a look at Fig. 43 for $n=13$. Non-obvious pairs are for the following vertices:
$$\begin{aligned} 3 &= \frac{1}{2}(-1+7) \equiv \frac{1}{2}(12+7), \\ 6 &= \frac{1}{2}(9+3), \\ 4 &= \frac{1}{2}(10-2) \equiv \frac{1}{2}(10+11), \\ 5 &= \frac{1}{2}(10+0) \equiv \frac{1}{2}(10+13). \end{aligned}$$

Have a look at Fig. 44 for $n=11$. Black and white disks are located symmetrically thus it is enough to indicate pairs for one of the colors.

$$1 = \frac{1}{2}(5-3) = \frac{1}{2}(5+8), \quad 3 = \frac{1}{2}(1+5), \quad 8 = \frac{1}{2}(2+14) = \frac{1}{2}(2+3).$$

There is a distinct gray disk, therefore for number $x$ the following equation $x = \frac{1}{2}(M+m)$ has to be satisfied, for instance, $x = 2$, $m = 1$, $M = 3$.

It is impossible to satisfy conditions for $n=7$. Assume there are no more minimal than maximal. Then there are no more than 3 of them. It is clear that there cannot be 2 of them. Same goes for their location, since in case they are consecutive, then the ones on the sides do not have a pair, in case two are adjacent, then each of them does not have a pair. In case they are all distinct, we have an alternating placement and two of them do not have a pair.

Therefore, the answer is: for all $n \ne 7$.