35. Let $n$ be a positive integer divisible by 4. Find the number of permutations $\sigma$ of $\{1,2,3, \cdots, n\}$ such that $\{\sigma(1), \sigma(2), \cdots, \sigma(n)\}$ is a permutation of $\{1,2, \cdots, n\}$, and $\sigma(j)+\sigma^{-1}(j)=n+1$ for all $j \in\{1,2, \cdots, n\}$, where $\sigma^{-1}(j)=i$ when $\sigma(i)=j$.
35. Let $n=4k$, if there exists some $j$ such that $\sigma(j)=j$, then $\sigma^{-1}(j)=j$, thus $n+1=\sigma(j)+\sigma^{-1}(j)=2j$. Since 4 divides $n$, this is impossible. Therefore, for any $i$, $\sigma(i) \neq i$. Assuming $\sigma(a)=b$, then $\sigma^{-1}(a)=n+1-b$, i.e., $\sigma(n+1-b)=a$. Also, $\sigma(b)=n+1-\sigma^{-1}(b)=n+1-a$, i.e., $\sigma^{-1}(n+1-a)=b$, so $\sigma(n+1-a)=n+1-\sigma^{-1}(n+1-a)=n+1-b$. Thus, we obtain a cycle of period 4 $(a, b, n+1-a, n+1-b)$, and it is easy to prove that these 4 numbers are distinct. Therefore, any such permutation $\sigma$ consists of $\frac{n}{4}=k$ cycles of period 4. Note that $b$ and $n+1-b$ can be swapped, thus yielding another cycle of period 4 $(a, n+1-b, n+1-a, b)$. Dividing $n=4k$ numbers into $2k$ pairs $\{j, n+1-j\}(1 \leqslant j \leqslant 2k)$, there is only 1 way to do this. Then, combining $2k$ pairs into pairs of four, there are $\frac{1}{k!} C_{2k}^{2} C_{2k-2}^{2} \cdots C_{2}^{2}=\frac{(2k)!}{2^k \cdot k!}$ ways. Here, dividing by $k!$ is because the order of selecting the pairs of four does not matter, and each pair of four corresponds to 2 permutations, and there are $k$ pairs of four, so the number of permutations $\sigma$ is $\frac{(2k)!}{2^k \cdot k!} \cdot 2^k=\frac{(2k)!}{k!}$.