It seems like something that should be pretty obvious but I don't quite get why would it be true. For example, in the case of 2-fold perfect numbers, or simply perfect numbers, it is evident because
$$\sigma(n) - n = n$$
so you just add all the divisors except itself.
We call a positive integer $n>1$ multi-perfect , if $k:=\frac{\sigma(n)}{n}$ is an integer , in the case $k=2$ the number is called perfect.
What about the multi-perfect numbers that are not perfect (the case $k>2$) ? Are they all pseudoperfect ?
Here I consider the case of $k$-perfect numbers where $k \geq 3$. (The case $k=2$ was considered in the OP.)
It has been conjectured that $k$-perfect numbers ought to be divisible by $k!$. From Peter's comment, this then implies that $k$-perfect numbers are also pseudoperfect. (Note that the conjecture mentioned also implies that there are no odd $k$-perfect numbers for $k \geq 2$.)
Update: As correctly pointed out by Peter, $459818240$ is $3$-perfect but not divisible by $6$. It is however, divisible by $20$, which still makes it pseudoperfect, as had already been pointed by Peter in a comment underneath the OP.