For a given $k \in \{1, 2, \dots, 21\}$, consider the number of permutations $N_k$ that satisfy the conditions such that $a_{21} = k$.
When $k \in \{1, 2, \dots, 11\}$, for $i = 1, 2, \dots, k-1$, there exist $a_{2i-1}, a_{2i}$ that are permutations of $k-i, k+i$ (if $k=1$, there exists no such $i$), and $a_{2j} = j+1$ ($2k-1 \le j \le 20$) (if $k=11$, there exists no such $j$), so $N_k = 2^{k-1}$.

Similarly, when $k \in \{12, 13, \dots, 21\}$, there is $N_k = 2^{21-k}$.
Therefore, the number of such permutations satisfying the condition is
$$$$
$\sum_{k=1}^{21}$ N_k = $\sum_{k=1}^{11} 2^{k-1} + \sum_{k=12}^{21} 2^{21-k} = (2^{11}-1) + (2^{10}-1)$ = 3070. $\quad \square$