By the "sum of divisors" function I mean the function $\sigma (n)= \sum_{d|n} d$.
If we choose $k=1$ then it is not possible that we have $\sigma (n)=n$ because $n$ always has at least two divisors, which are $1$ and $n$, so we must have $\sigma (n) \geq n+1$.
But if we choose $k=2$ then we can have $\sigma (n)=2n$ because perfect numbers satisfy this equality and they exist.
So the natural thing to ask is:
Is it true that for every $k \in \mathbb N$; $k \geq 2$ there exist at least one $n(k) \in \mathbb N$ such that we have $\sigma (n(k))=k \cdot n(k)$ (I write here $n(k)$ to denote dependence of $n$ on $k$)?
An integer $n$ is called (multi)-$k$-perfect if $\sigma(n)=kn$. So the question is, whether or not for every $k\ge 2$ there exists a $k$-perfect number. This is certainly true for small $k$. For references see, for example, here, and also here.
So it seems to be an open problem, but Erdös has conjectured that it is true for every $k\ge 2$.