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

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 $n\geq 3$.

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