Find a vector NOT perpendicular to a given set of vectors 
Let us suppose to have a finite set of vectors $S=\{v_1,\ldots,v_m\}$ in $\mathbb{R}^n$ (with $m \gg n$ in general). I need to find a vector $x \in \mathbb{R}^n$ that is NOT perpendicular to any vector in $S$. The existence of such a vector $x$ is guaranteed, moreover almost every vector in $\mathbb{R}^n$ satisfies this property. But I need to find an algorithm to determine a vector with these properties, I cannot close my eyes and choose.

Any ideas?
EDIT1: in my case, the vectors in $S$ have some "symmetries" in the sense that they are generated by permutation and change of signs of a few vectors in $S$
EDIT2: $v_1+\ldots+v_m=0 \in \mathbb{R}^n$
EDIT3: I simplified the solution proposed by Tom Collinge. We define a sequence $\{x_i\}_{i=1}^m$ such that $x_m=x$ is what we are looking for. First define $x_1=v_1$, so $x_1 \cdot v_1 \ne 0$ and they are not perpendicular (obviously). Then, recursively, define for $k \in \{2,\ldots,m\}$
$$
x_k=\begin{cases}
x_{k-1}+2v_k & \text{if }x_{k-1}=-v_k\\
x_{k-1}+v_k & \text{otherwise}
\end{cases}.
$$
By construction we have that $x_k \cdot v_i \ne 0$ for $i \le k$, then $x_m \cdot v_i \ne 0$ for all $v_i \in S$, so $x=x_m$ is not perpendicular to every element in $S$. Can it works?

EDIT4: As pointed out, the algorithm proposed in EDIT3 does not work


If you can implement on a computer, perhaps the following would work.
Create an $m \times n$ matrix $A$, the rows of which contain your vectors.
Now pick a random uniform vector $b > \vec{0} \in \mathbb{R}^m$, for example by generating the entrees of $\vec{b}$ uniformly, and solve $A \vec{x} = \vec{b}$.
If any solution is found, you are done. If not, re-generate $\vec{b}$ and do it again.
If your assertion that practically every vector in $\mathbb{R}^n$ will do is correct, you will get a viable answer in a very small amount of tries.
