How prove this can choose two postive integer numbers $a_{m},a_{k},$such $\frac{a_{m}+a_{k}}{3a_{p}}\notin N^{+},$ 
If  $a_{1},a_{2},\cdots,a_{n}(a_{i}\neq a_{j}),n\ge 3$ are positive integers,show that: we can always choose two positive integers among them, $a_{m},a_{k},m,k\in\{1,2,\cdots,n\}$.such that
  $$\dfrac{a_{m}+a_{k}}{3a_{p}}\notin N^{+},\forall p\in\{1,2,3,\cdots,n\}$$

This problem is from the Jiangxi province Mathematical Contest,2014 .(I failed This year's exam was very difficult)
my idea: if $n=3$,Assume that $a_{1}=1,a_{2}=2,a_{3}=3$,then we have $a_{p}=1$,then we can choose $a_{m}=2,a_{k}=3$,then it is 
clearly
$$\dfrac{a_{m}+a_{k}}{3a_{p}}=\dfrac{5}{3}\notin N^{+}$$
if $a_{p}=2$,then we can choose $a_{m}=1,a_{k}=3$
if $a_{p}=3$  then we can choose $a_{m}=1,a_{k}=2$,
But for in general $n$,I can't prove it.

Thank you very much
 A: We can assume $a_1 < a_2 < a_3 < \ldots <a_n$ without loss. Since
$n\geq 3$, we have $a_n \geq 3$.
Let us argue by induction on $m=a_n$. When $m=3$, we have
$n=3, (a_1,a_2,a_3)=(1,2,3)$ and this case has already been
treated by the OP.
Now, suppose that $m>3$ and that the result is true for $m-1$.
Let $a_1 < a_2 < a_3 < \ldots <a_n$ be a sequence of positive integers.
If, for some $i \neq j$, $a_i+a_j$ is not divisible by $3$, then we may
take $m=i,k=j$ and we are done. 
So we can assume that all the $a_i+a_j(i \neq j)$ are divisible by $3$.
So for any distinct $i,j,k$, we have that $a_j-a_i=(a_j+a_k)-(a_i+a_k)$ is divisible by
$3$. So all the $a_i$ are congruent to a constant 
$a\in\lbrace 0,1,2\rbrace$ modulo $3$. Then $2a$ is divsible by $3$,
so $a=0$. We see then that all the $a_i$ are divisible by $3$ ; write
$a_i=3b_i$ where $b_i$ is a positive integer. By the induction hypothesis,
there are indices $m\neq k$ such that $\frac{b_n+b_m}{3b_p} \not\in {\mathbb N}$
for any $p$. Since $\frac{a_n+a_m}{3a_p}=\frac{b_n+b_m}{3b_p}$, we are done.
