Let $m\geq 1$ an integer and $\sigma(m)=\sum_{d\mid m}d$ the sum of divisors function, then Möbius inversion formula states $$m=\sum_{d\mid m}\sigma(d)\mu\left(\frac{m}{d}\right),$$ since this convolution is commutative.
If we assume that $n=\prod_{p\mid n}p^{e_p}$ is a perfect number, that is an integer such that $\sigma(n)=2n$, (an example is $6=2^1\cdot 3^1$) then we can write that $$\sigma(n)=2\sum_{d\mid n}\sigma(d)\mu\left(\frac{n}{d}\right)$$ holds, and because the sum of divisor function is a multiplicative arithmetic function then $$\prod_{p\mid n}\sigma\left(p^{e_p}\right)=\sigma(n)=2\sum_{d\mid n}\sigma(d)\mu\left(\frac{n}{d}\right)$$ (where I've written LHS in such expanded way with the purpose to show it, if you know from this identity how get more interesting facts than the following mine). In particular dividing by $\sigma(n)=2n$ it is obvious then that $$\left(2\sum_{\substack{d\mid n\\d<n}}\sigma(d)\mu\left(\frac{n}{d}\right)\right)+0\equiv 0\operatorname{mod}2n,$$ and if we presume that $n$ also is odd, thus we've in our hands an odd perfect number $n$ then the simplification rule for congruences states $$\sum_{\substack{d\mid n\\d<n}}\sigma(d)\mu\left(\frac{n}{d}\right)\equiv 0\operatorname{mod}n.$$
Question. Please can do you argue if the following calculations are rights and if the fact that I've claimed for odd perfect numbers is obvious? Thus I am asking if do you know if there were some mistakes in previous simple calculations, and secondly if is known the sequence of odd integers $t>1$ satisfying $$\sum_{\substack{d\mid t\\d<t}}\sigma(d)\mu\left(\frac{t}{d}\right)\equiv 0\operatorname{mod}t.$$ If it's possible what are the first terms $t_k$ of such sequence. I believe that answering it, from my viewpoint, could to tell us if there is something interesting in my calculations. Many thanks.