Proving a Property of a Set of Positive Integers I have a question as such:

A set $\{a_1, \ldots , a_n \}$ of positive integers is nice iff
  there are no non-trivial (i.e. those in which at least one component
  is different from $0$) solutions to the equation $$a_1x_1 + \ldots +
> a_nx_n = 0$$ with $x_1 \ldots x_n \in \{ -1, 0, 1 \}$. Prove that any
  nice set with $n$ elements necessarily contains at least one element
  that is $\geq \frac{2^n}{n}$.

Here's my work so far:
Let $A = \{a_1, \ldots, a_n \}$. Let $P = a_1x_1 + \ldots + a_nx_n$.  $A$ is nice iff there are only trivial solutions to $P$.
There is a nontrivial solution to $P$ iff $\exists B,C \subset A$ s.t. $sum(B)=sum(C)$. Note that $B\cap C = \emptyset$, but $B\cup C$ does not necessarily equal $A$, as once there are such subsets,
the other elements of $A$ can be set to multiply with $x_i = 0$ s.t. Moreover, either $B$ or $C$ has to be multiplied by $x_i = -1$ s.t. $B+C = 0$. 
The contrapositive of the initial statement is that if $\forall a \in A, a < \frac{2^n}{n}$, then $A$ is not nice. 
We note that a non-nice set can contain $x \geq \frac{2^n}{n}$. It is only that if a set is nice that it necessarily contains $x \geq \frac{2^n}{n}$. Just the fact that a set contains such an element doesn't actually tell us whether the set is nice or not.
The question that I arrived at is this: why is that if $\forall a\in A, a < \frac{2^n}{n}$, then $A$ has a non-trivial solution? If I can show why this is true, it would constitute proof of the contrapositive, and I would be done.
By the way, if anyone can think of a better title; please do suggest it.

Hint: If the sums of two distinct subsets are equal, then the set is not nice.
There are $2^n$ subsets. If the numbers are all $\lt \frac{2^n}{n}$, then the sum of all the numbers is less than $2^n$. Now use the Pigeonhole Principle. 
