Find 10 distinct positive integers such that each of them divides the sum of these 10 integers

I started by finding a simplier case, with just a set of 3 numbers:

$1 + 2 + 3 = 6$

Now I want to add another number $n$ to this set, but making sure that $6+n$ is still divsiable by $1, 2, 3$ and $n$:

$n = 4:\\ 1 + 2 + 3 + 4= 10 \rightarrow 3\nmid10$ 

$n = 5:\\ 1 + 2 + 3 + 5 = 11 \rightarrow 2,3,5\nmid11$

$n = 6:$
*$1 + 2 + 3 + 6 = 12$*

Notice that we just added the sum of our previous set and got a new set, which still satisfies our requirements. Adding 12 to our new set yields a newer set:

$1 + 2 + 3 + 6 + 12 = 24$

And again this newer set is satisfactory. I can then repeat this process until I have 10 distinct numbers:

$1+2+3+6+12+24+48+96+192+384=768$
This method can be generalized to get a set of $N$ positive integers:

 $S = \{d_1, d_2, d_3,..., d_{k}, p, 2p, 4p, 8p, ..., 2^{N-k-1}p\}$

Where $d_1 + d_2 + d_3 + ... + d_k = p$ and $N > k$.
This is chiefly in the realm of "psuedoperfect numbers". A number is considered "perfect" if its equal to the sum of its divsors (not including itself). 6 is the simplest perfect number, and they exponentially increase from there (https://oeis.org/A000396). Meanwhile, a number is considered "psuedoperfect" if its equal to the sum of not of its divisors.