What is the smallest number $n>2$ of this kind? In defining perfect numbers we want that the sum of its divisors (including itself) is equal twice the number itself. We can define more general classes of numbers such that sum of its divisors is divisible by the number itself, i.e. we can ask is it true that for every $k \in \mathbb N (k>1)$ there exists at least one (or infinitely many) natural numbers such that the sum of their divisors is equal to $kn$.
But what happens if we sum non-divisors?
As an example let us take the number $12$. It is divisible by $1,2,3,4,6,12$ so the sum of its non-divisors is $5+7+8+9+10+11=50$ and $50$ is not a multiple of $12$.
What is the smallest number $n>2$ such that the sum of its non-divisors is a multiple of $n$?
(if you find more than one number of this kind feel free to post it in an answer)
 A: Let $n$'s factorization be $\prod_{i=1}^kp_i^{k_i}$. Then we get $\sigma(n)=n\prod_{i=1}^k(1+1/p_i+\cdots+1/p_i^{k_i})$. Therefore, the sum of non-divisors is$$\frac{n(n+1)}{2}-n\prod_{i=1}^k\left(1+\frac{1}{p_i}+\cdots+\frac{1}{p_i^{k_i}}\right)$$
and we are looking for $n$ which makes this number multiple of $n$. By dividing by $n$, the condition becomes$$\frac{n+1}{2}-\prod_{i=1}^k\left(1+\frac{1}{p_i}+\cdots+\frac{1}{p_i^{k_i}}\right)$$is integer.
If $n$ is odd, then the problem becomes finding odd $n$ which makes $\prod_{i=1}^k(1+1/p_i+\cdots+1/p_i^{k_i})$ integer. Such $n$ will have integral abundancy, and will be odd multiperfect number! No such number was discovered so far, and their non-existence was not proved.
If $n$ is even, there are many numbers known to have "half-integral" abundancy. You can check the list at A159907.
A: Number 24
Here is a small ruby script
for a in 3...100
  mysum = 0
  for i in 1...a
    if a % i != 0 
      mysum += i
    end
  end
  if mysum % a == 0
    p "I found number #{a}"
    break;
  end
end

