Try to construct this before solving it

Let the parabola $P_{1}$ be represented by $x^2 = 4py$ (with directrix $l_{1}: y = -p$ and focus $F_{1}(0,p)).$ If the focus $F_{2}(1,p)$ lies on $P_{1}$ , then we obtain: $1 = 4p^2 \Rightarrow p =\frac{1}{2}$ , or $\boxed{x^2=2y}$ for $P_{1}.$

For the parabola $P_{2}$ , it must be concave down in order to satisfy $l_{2} \parallel l_{1} \parallel F_{1} F_{2}$ and $l_{2} \neq l_{1}.$ This requires the form: $-(x-1)^2 = 4q[y-(1/2+q)]$ , and if $F_{1}(0,1/2) \in P_{2}$ then we obtain:

$-(0-1)^2 = 4q[1/2 - (1/2+q)] \Rightarrow -1 = -4q^2 \Rightarrow q = \frac{1}{2}$

or $\boxed{-(x-1)^2 = 2(y-1)}$ for $P_{2}.$ We now turn to solving for the intersection points $A$ and $B$ , which compute to:

$-(x-1)^2 = 2y-2 \Rightarrow -x^2+2x-1 = x^2-2 \Rightarrow 0 = 2x^2 -2x -1 \Rightarrow x = \frac{2 \pm \sqrt{4-4(2)(-1)}}{4} \Rightarrow \boxed{x = \frac{1 \pm \sqrt{3}}{2}}.$

and $y = \frac{x^2}{2} \Rightarrow \boxed{y = \frac{1}{2} \pm \frac{\sqrt{3}}{4}}.$ Thus, we have the points $A(\frac{1+\sqrt{3}}{2}, \frac{1}{2}+\frac{\sqrt{3}}{4})$ and $B(\frac{1-\sqrt{3}}{2}, \frac{1}{2}-\frac{\sqrt{3}}{4})$ , and:

$AB^{2} =[\frac{1+\sqrt{3}}{2} - \frac{1-\sqrt{3}}{2}]^2 + [\frac{1}{2}+\frac{\sqrt{3}}{4} - \frac{1}{2} +\frac{\sqrt{3}}{4}]^2 = (\sqrt{3})^2 + (\frac{\sqrt{3}}{2})^2 = 3 + \frac{3}{4} = \boxed{\frac{15}{4}} = \frac{m}{n}$

Finally, we obtain $1000m+n = 1000(15) + 4 = \boxed{1504}.$