Positive finite sum with restriction Is it possible to satisfy the condition $\sum_{i=1}^N \sum_{j=1}^N\frac{q_i q_j}{a_{ij}}>0$, given that $q_i$ and $q_j$ are non-zero integers, $a_{ij}$ are positive real numbers and $\sum_{i=1}^N q_i =0$? As a special case, allow $N \rightarrow \infty$.
Another formulation of the problem: For a set of integers $A=\left(q_1,q_2,...q_N\right)$, such that $\sum_{i=1}^N q_i =0$, is it possible to find positive weights $a_{ij}$ such that $\sum_{i=1}^N \sum_{j=1}^N\frac{q_i q_j}{a_{ij}}>0$?
Apologies for the edit, I didn't formulate the problem very well earlier. 
It is probable I am missing a simple proof here, but I can't seem to find it. Thanks!
 A: Let $M$ denote the matrix whose entries are $M_{ij} = \frac{1}{a_{ij}}$.  Let $q$ denote the column vector $(q_1,\dots,q_N)$. Then we have
$$
\sum_{i=1}^N \sum_{j=1}^N\frac{q_i q_j}{a_{ij}} = q^TMq
$$
with that in mind: if $M$ is positive definite, then $q^TMq$ will be positive whenever $q \neq (0,\dots,0)$.
For example, if we take 
$$
a_{ij} = 
\begin{cases}
1 & i=j\\
2 & \text{otherwise}
\end{cases}
$$
then we will have $\sum_{i=1}^N \sum_{j=1}^N\frac{q_i q_j}{a_{ij}} > 0$ whenever $q$ are not all equal to $0$.

A criterion that should help in the infinite dimensional case: a sufficient condition to make this work is to have $a_{ij} = a_{ji}$ for all $i,j$, and
$$
\frac{2}{a_{ii}} > \sum_{j= 1}^N \frac{1}{a_{ij}} \qquad \forall i=1,\dots,N
$$
which is to say that
$$
a_{ii} < 2\left(\sum_{j= 1}^N \frac{1}{a_{ij}}\right)^{-1}\qquad \forall i
$$
Or if you prefer
$$
a_{ii} < \left(\sum_{j\neq i} \frac{1}{a_{ij}}\right)^{-1} \qquad \forall i
$$
This condition will continue to be sufficient as $N \to \infty$.
