One more motivated by recent questions of [Zhi-Wei Sun](https://mathoverflow.net/users/124654/zhi-wei-sun).

Let $S_n$ be the group of permutations of $\{1,2,\ldots, n\}$.

Is it true that, for every $n \ge 8$, there is at least one even permutation $\pi \in S_n$ and at least one odd permutation $\tau \in S_n$ with $$\sum_{k=1}^n \frac{1}{k \, \pi(k)} = \sum_{k=1}^n \frac{1}{k \, \tau(k)} = 1?$$

One case for each $n$ is not hard; I made it a math.stackexchange [question](https://math.stackexchange.com/questions/3013470/permutations-products-and-unit-fractions) that was successfully answered in 12 minutes.  Hopefully the other case is more interesting.  

Here are the numbers of even and odd permutations satisfying the sum condition for small $n$.

\begin{array}{c|rr}
n\backslash \text{sgn} & +1 & -1 \\ \hline
1 & 1 & 0 \\
2 & 0 & 1 \\
3 & 2 & 0 \\
4 & 0 & 2 \\
5 & 4 & 0 \\
6 & 0 & 2 \\
7 & 4 & 0\\
8 & 6 & 4\\
9 & 12 & 24\\
10 & 90 & 88
\end{array}

One of the first ``interesting'' permutations is the even permutation (in cycle notation) $(1,2,5,8,7,6)(3,4) \in S_8$ which gives
\begin{align*}
\frac{1}{1\cdot2} + \frac{1}{2\cdot5}+ \frac{1}{3\cdot4}+ \frac{1}{4\cdot3}+ \frac{1}{5\cdot8}+ \frac{1}{6\cdot1}+ \frac{1}{7\cdot6}+ \frac{1}{8\cdot7}\\
= \frac{1}{2} + \frac{1}{10}+ \frac{1}{12}+ \frac{1}{12}+ \frac{1}{40}+ \frac{1}{6}+ \frac{1}{42}+ \frac{1}{56}=1.
\end{align*}
Not coincidentally, $n=8$ is the smallest value for which there are non-$n$-cycle permutations that satisfy the sum condition.