I think it's $11$. $30030=2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13$. We regard the six prime divisors as the vertices of a graph, and turn the problem into the following, more general one: Find the minimal $e=e(t)$ with the property that every graph on $2t$ vertices with $e$ edges has a perfect matching (in our case, $2t=6$). The answer should be $\frac{(2t-1)(2t-2)}2+1$ (which gives $11$ when $2t=6$), I believe. We can prove this as follows: $\frac{(2t-1)(2t-2)}2$ certainly isn't enough, since we can take a complete graph on $2t-1$ vertices, and an isolated vertex. This graph obviously has no perfect matching. On the other hand, $\frac{(2t-1)(2t-2)}2+1$ is enough, since it has been shown here that any graph on $u$ vertices with $\frac{(u-1)(u-2)}2+2$ edges is hamiltonian. We apply his to $u=2t$ to conclude that our graph, which has $2t$ vertices but only $\frac{(2t-1)(2t-2)}2+1$ edges, must have a hamiltonian path $x_1,x_2,\ldots,x_{2t}$. Our perfect matching will then consist of the pairs of vertices $(x_i,x_{i+1}),\ i\in\{1,3,\ldots,2t-1\}$.

I'm sorry to ask, Grobber, but how is relation between two vertices defind?

Two vertices=prime numbers are connected iff their product occurs in the set of that $n$ numbers.