Show that a number $n$ is divisible by 6 if and only if it can be written as a sum of three distinct divisors. If $6|n$ then $n=6k=3k+2k+k$. And $3k|n$, $2k|n$ and $k|n$.
Now let $p,q$ and $r$ be three distinct divisors of $n$ so that :
$$n=p+q+r$$
Because $p|n $, $ q|n$ and $r|n$ I figured that $p|q+r $, $ q|p+r$ and $r|p+q$.
I tried to prove that n is even or a multiple of 3 but without much luck.
How can I prove the statement?

Without, loss of generality, presume that $p > q > r$. Observe that $$p|(q+r) \implies (q+r) = k_1 p \implies 1 \le k_1 = \frac{q+r}{p} < 2 \implies k_1 =1.$$ 
As such, $p = q+r$, and $n = 2(q+r)$.
Since $q|n=2(q+r)$, we have $q|2r$. That is, $$2r = k_2 q \implies 1 \le k_2=\frac{2r}{q}<2 \implies k_2 = 1.$$
Therefore, $$q = 2r \implies p = q+r =3r \implies n = 6r.$$
