Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?

Question

Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?
I'm not really sure how to even approach this question.
Source: Washington's Monthly Math Hour, 2014

Answer 1

Hint: $2,4,6$ are all divisors of $2+4+6=12$. Similarly $2+4+6+12$ are all divisors of $2+4+6+12=24$

So too are $2,4,6,12,24$ all divisors of $2+4+6+12+24$

$~$

Claim: Let $x_1=2, x_2=4, x_3=6$ and let $x_{n+1} = \sum\limits_{k=1}^n x_k$ for each $n\geq 3$. You have that $x_i\mid \sum\limits_{k=1}^nx_k$ for all $i\leq n$ for all $n\geq 3$.

Answer 2

$$\begin{align} 1+2+3&=6\\ 1+2+3+6&=12\\ 1+2+3+6+12&=24\\ \vdots \end{align}$$