Suppose that exactly $x$ of the lines are parallel to each other (so there must be no intersections in within these lines). Then there are $56 - x$ lines remaining, and they intersect with each other at most $\binom {56-x}{2}$ times, or $\frac{(56-x)(55-x)}{2}$ in this case. Now each of these lines can intersect the parallel lines at most $x$ times, so there are at most $(56 - x)x$ intersections in this case. So we need to solve for $x$ in the equation $$\frac{(56-x)(55-x)}{2} + (56 - x)x \geq 594$$ Multiplying both sides by two and using the distributive property we get $$(56-x)(55+x) \geq 1188 = 12 \cdot 99$$ So we find that $x = \boxed{44}$ satisfies the inequality perfectly here and any higher value of $x$ will clearly fail