Let $S_n$ be the symmetric group of all the permutations of $\{1,\ldots,n\}$. Recall that a permutation $\sigma\in S_n$ is called a derangemnt if $\sigma(k)\not=k$ for all $k=1,\ldots,n$.

Motivated by the well-known result that $\sum_{k=m}^n\frac1k\not\in\mathbb Z$ whenever $n\ge m>1$ (Kurschak, 1918), here I ask the following question.

QUESTION: Is it true that whenever $n\ge m\ge1$ we have $$\sum_{k=m}^n\frac{\sigma(k)}k\not\in\mathbb Z$$ for all derangements $\sigma\in S_n$?

If $n$ is a prime number $p$ and $\sum_{k=m}^n\frac{\sigma(k)}k\in\mathbb Z$ with $\sigma\in S_n$, then $\sigma$ is not a derangement since $\sigma(p)=p$. Thus the question has a positive answer if $n$ is prime. I conjecture that the question always has a positive answer, and I have verified this for every $n=1,\ldots,11$. For $n=4$, note that $$\frac41+\frac12+\frac33+\frac24\in\mathbb Z$$ but the permutation $(4,1,3,2)$ of $\{1,2,3,4\}$ is not a derangement since it fixes the number $3$.