I will prove this cool result by inducting on $n$. Verify base cases manually, and assume the hypothesis holds, up to $n-1$. Now, we analyze two cases, depending on the parity of $n$: Let $n=2\ell$. Note that, $p(k)=k$ if $k$ is odd, and $p(2\ell)=p(\ell)$. With this, $$ \sum_{k=1}^n \frac{p(k)}{k} = \ell + \sum_{k=1}^{\ell}\frac{p(2\ell)}{2\ell}=\ell+\frac12 \sum_{k=1}^\ell \frac{p(\ell)}{\ell}. $$Now, by the inductive hypothesis, it holds that $\sum_{k=1}^\ell \frac{p(\ell)}{\ell} \in (2\ell/3 , 2(\ell+1)/3)$. Consequently, $$ \frac{2n}{3}=\frac{4\ell}{3} < \sum_{k=1}^n \frac{p(k)}{k} < \frac{4\ell+1}{3}=\frac{2n+1}{3}<\frac{2(n+1)}{3}, $$as requested. The case where $n=2\ell+1$ is handled analogously. This concludes the proof.