Rigorous proof of $n\sum_{d\mid n}\frac{1}{d}=\sum_{d\mid n}d$ for $n\in\mathbb N$.

Question

Let $n\in\mathbb N$ and consider the sum

$$\sum_{d\mid n}\frac{n}{d}.$$

I understand that dividing $n$ by any of it's divisors is also a divisor of $n$ so intuitively it would make sense that $\{\frac n d\ :\ d\mid n \}=\{d\ :\ d\mid n\}$, from which the desired equality in the title is acquired. but I cannot "see" how this can be proven rigorously. Any help or insight would be appreciated!

Edit: In many sources, including the question this one is marked as a possible duplicate to, it is not explained why the identity holds, it's just used. If it actually is explained, it's not clear to me, at least. I am interested in an actual explanation, that's why my question differs from the one marked as duplicate.

Answer 1

Exactly the right idea, noting that $d \mid n \iff \frac{n}{d} \mid n$. $$\begin{align} n \sum_{d \mid n} \frac{1}{d} &= \sum_{d \mid n} \frac{n}{d} \\ &= \sum_{\frac{n}{d} \mid n} \frac{n}{d} \\ &= \sum_{m \mid n} m \\ &= \sum_{d \mid n} d \end{align}$$