$p\nmid 2n-1,$ then $\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}\equiv 0 \pmod{p^3} \Leftrightarrow \sum_{k=1}^{p-1}\frac{1}{k^{2n}}\equiv 0 \pmod{p^2} $ Is it true that if $p$ is a prime and $p\nmid 2n-1,$ then
$$\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}\equiv 0 \pmod{p^3} 
 \hspace{12pt}\Leftrightarrow  \hspace{12pt}
 \sum_{k=1}^{p-1}\frac{1}{k^{2n}}\equiv 0 \pmod{p^2} $$
I find some examples: $(2n-1,p)=(3,37)(7,67)(7,877)(9,5)(13,7)(13,59)(13,607)$
 A: We show something stronger, namely that
$$
2\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}+(2n-1)\cdot p\cdot\sum_{k=1}^{p-1}\frac{1}{k^{2n}}\equiv 0\mod{p^3}.
$$
Let $n\geq 1$ be a fixed integer. If we compute the difference
$$
\frac{1}{(1+x)^{2n-1}}-\left(1-(2n-1)x+n(2n-1)x^2\right),
$$
we see that it is of the form
$$
\frac{f(x)}{(1+x)^{2n-1}},
$$
where $f(x)\in\mathbb{Z}[x]$ is some polynomial which is divisible by $x^3$. It follows that for any $x\in p\mathbb{Z}_{(p)}$,
$$
(\star)\,\,\,\,\,\,\,\,\,\,\frac{1}{(1+x)^{2n-1}}\equiv 1-(2n-1)x+n(2n-1)x^2\mod{p^3}.
$$
Now, we have
$$
\begin{eqnarray*}
\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}&=&\sum_{k=1}^{p-1}\frac{1}{(p-k)^{2n-1}}\\
&=&-\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}\frac{1}{\left(1-\frac{p}{k}\right)^{2n-1}}\\
&\equiv&-\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}-(2n-1)p\sum_{k=1}^{p-1}\frac{1}{k^{2n}}-n(2n-1)p^2\sum_{k=1}^{p-1}\frac{1}{k^{2n+1}}\mod{p^3},
\end{eqnarray*}
$$
where in the third line we have used $(\star)$ with $x=-\frac{p}{k}$.
For $p\geq 2n+3$,
$$
\sum_{k=1}^{p-1}\frac{1}{k^{2n+1}}\equiv 0\mod{p}
$$
(perhaps you are familiar with this fact?), so in this case we have
$$
2\sum_{k=1}^{p-1}\frac{1}{k^{2n-1}}+(2n-1)\cdot p\cdot\sum_{k=1}^{p-1}\frac{1}{k^{2n}}\equiv 0\mod{p^3},
$$
implying the fact you want.
