Permutations $\pi\in S_n$ with $\sum_{k=1}^n\frac1{k+\pi(k)}=1$ Let $S_n$ be the symmetric group of all the permutations of $\{1,\ldots,n\}$.
Motivated by Question 315568 (http://mathoverflow.net/questions/315568), here I pose the following question.

QUESTION: Is it true that for each integer $n>5$ we have $$\sum_{k=1}^n\frac1{k+\pi(k)}=1$$ for some odd (or even) permutation $\pi\in S_n$? 

Let $a_n$ be the number of all permutations $\pi\in S_n$ with $\sum_{k=1}^n(k+\pi(k))^{-1}=1$. Via Mathematica, I find that
\begin{gather}a_1=a_2=a_3=a_5=0,\ a_4=1,\ a_6=7,
\\ a_7=6,\ a_8=30,\ a_9=110, \ a_{10}=278,\ a_{11}=1332.\end{gather}
For example, $(1,4,3,2)$ is the unique (odd) permutation in $S_4$ meeting our requirement for $n=4$; in fact, 
$$\frac1{1+1}+\frac1{2+4}+\frac1{3+3}+\frac1{4+2}=1.$$
For $n=11$, we may take the odd permutation $(4,8,9,11,10,6,5,7,3,2,1)$ since
\begin{align}&\frac1{1+4}+\frac1{2+8}+\frac1{3+9}+\frac1{4+11}+\frac1{5+10}\\&+\frac1{6+6}+\frac1{7+5}+\frac1{8+7}+\frac1{9+3}+\frac1{10+2}+\frac1{11+1}\end{align} has the value $1$, we may also take the even permutation $(5, 6, 7, 11, 10, 4, 9, 8, 3, 2, 1)$ to meet the requirement.
I conjecture that the question has a positive answer. Your comments are welcome！
PS: After my initial posting of this question, Brian Hopkins pointed out that A073112($n$) on OEIS gives the number of permutations $p\in S_n$ with $\sum_{k=1}^n\frac1{k+p(k)}\in\mathbb Z$, but A073112 contains no comment or conjecture.
 A: Claim: $a_n>0$ for all $n\geq 6\quad (*)$.  
Proof: We use induction to prove $(*)$.
We have $a_6,a_7,a_8,a_9,a_{10},a_{11}>0$. Assume $(*)$ holds for all the integers $\in [6,n-1]$. We want to show that $a_n>0$ for all $n\geq12$.
If $n$ is an odd, let $n=2m+1$, we have $m\geq6$, so by induction hypothesis, there exists $\pi\in S_m$ such that $\sum\limits_{k=1}^{m}\frac{1}{k+\pi(k)}=1$. 
Let
\begin{align*}
\sigma(2k+1)&=2m+1-2k\quad\text{for}\quad k=0,1,2\ldots, m,\\
\sigma(2k)&=2\pi(k)\quad\text{for}\quad k=1,2,\ldots, m.
\end{align*}
Then $\sigma\in S_{n}$, and 
$$\sum\limits_{k=1}^{n}\frac{1}{k+\sigma(k)}=\sum\limits_{k=1}^{m}\frac{1}{2k+\sigma(2k)}+\sum\limits_{k=0}^{m}\frac{1}{2k+1+\sigma(2k+1)}\\
=\frac{1}{2}\sum\limits_{k=1}^{m}\frac{1}{k+\pi(k)}+\sum\limits_{k=0}^{m}\frac{1}{2k+1+(2m-2k+1)}=\frac{1}{2}+(m+1)\frac{1}{2m+2}=1.$$
If $n$ is an even, let $n=2m$, also we have $m\geq6$ and there exists $\pi\in S_m$, such that $\sum\limits_{k=1}^{m}\frac{1}{k+\pi(k)}=1$. 
Let
\begin{align*}
\sigma(2k-1)&=2m+1-2k\quad\text{for}\quad k=1,2,\ldots,m, \\
\sigma(2k)&=2\pi(k)\quad\text{for}\quad k=1,2,\ldots,m.
\end{align*}
Then $\sigma\in S_{n}$ and 
$$\sum\limits_{k=1}^{n}\frac{1}{k+\sigma(k)}=\sum\limits_{k=1}^{m}\frac{1}{2k+\sigma(2k)}+\sum\limits_{k=1}^{m}\frac{1}{2k-1+\sigma(2k-1)}=\frac{1}{2}\sum\limits_{k=1}^{m}\frac{1}{k+\pi(k)}+
\sum\limits_{k=1}^{m}\frac{1}{2k-1+(2m-2k+1)}=1.$$
Hence $a_{n}>0$.
By induction $(*)$ holds for all the $n\geq 6$.
