I don't suppose it would something like :

2C1 groups 2C1 b or girl 2n!/n! 2n!/n! all that would be the #ways you can get and equal # of boys and girls in each grouo: 2*2*(2n!/n!)^2

Then the total # ways to divide them would be

4nC2n * 2nC2n?

so the probability would be

4*[(2n)!/n!]^2 / 4nC2n

meh I'm just making this up kinda so XP