Solution:

The answer is no. We shall prove by induction that for any $k \geq 3$ there is a permutation $a_{1}, a_{2}, \ldots, a_{k}$ of $1,2, \ldots, k$ such that
$$a_{m}+a_{n} \neq 2 a_{\frac{m+n}{2}} \text{ for any } 1 \leq m<n \leq k \text{ of the same parity. }$$
For $k=3$ and $k=4$ take the permutations $1,3,2$ and $1,3,2,4$, respectively. Assume that our statement is true for any integer less than $k$. Start with the following permutation of $1,2, \ldots, k$: the odd numbers are in the first block, the even numbers that are not divisible by $4$ are in the second block, etc. For example, if $k=12$, then we have $1,3,5,7,9,11 ; 2,6,10 ; 4,12 ; 8$.
If $a_{m}$ and $a_{n}$ are in different blocks, set $a_{m}=2^{s} b$ and $a_{n}=2^{t} c$, where $t>s>0$ and $b, c$ are odd integers. Then $\frac{a_{m}+a_{n}}{2}=2^{s-1}\left(b+2^{t-s} c\right)$ is in the block before that of $a_{m}$, whence $a_{\frac{m+n}{2}}$ is in a block between these of $a_{m}$ and $a_{n}$. So (1) holds.
It remains to reorder the integers in any block in such a way that (1) is satisfied for the numbers in this block. Consider the $(r+1)$-th block: $2^{r}, 3\cdot2^{r}, \ldots,(2 d-1) 2^{r}$, where $2 d-1 \leq k$. By the induction assumption there is a permutation $b_{1}, b_{2}, \ldots, b_{d}$ of $1,2, \ldots, d$ satisfying (1). Set $c_{i}=2 b_{i}-1$ for $i=1,2, \ldots, d$ and consider the permutation $2^{r} c_{1}, 2^{r} c_{2}, \ldots, 2^{r} c_{2 d-1}$. Then
$$\frac{c_{m}+c_{n}}{2}=b_{m}+b_{n}-1 \neq 2 b_{\frac{m+n}{2}}-1=c_{\frac{m+n}{2}}$$
which completes the proof.