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 reuirement.

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)in math.stackexchange.com/a/2597700 (which has since become part of Theorem 1.1 in Johann Cigler's arXiv:1803.05164v2). Indeed, if I shift my numbers $0, 1, \ldots, n-1$ by $1$, then my nimble permutation becomes a $\pi \in S_n$ such that each $i \in \left\{1,2,\ldots,n\right\}$ has the property that $i + \pi\left(i\right)$ equals a power of $2$ minus $1$. Are your and my permutation related? $\endgroup$ – darij grinberg Nov 27 '18 at 3:13