Well, if we have a random arrangement (boys indistinguishable, girls indisinguishable), such as:
BGGBGGBGGBGGBGGBGGBG (not terribly random, TBH, but it doesn't matter!)
then there are precisely 7!13! ways of doing this where you do distinguish the boys and girls. (7! ways of arranging the boys, 13! ways of arranging the girls).
So if you count the number of scenarios where the i_th position is BG or GB (i.e. the N_i of the question), then you can go from indistinguishable to distinguishable by multiplying by 7!13!.
But the total number of arrangements increases from 20C7 = 20!/(7!13!) to 20! as you go from indistinguishable to distinguishable, and so the average is unchanged.
Basically, what you're averaging doesn't care if the Boys are distinguisable, so in the end either method of counting gives the same result.
I feel it would be remiss to ignore the "probability" method of solving this question:
Consider the i, i+1 positions, define I_i = 1 if we have BG or GB in these positions. 0 otherswise. Now p(BG) + p(GB) = 2 * 7 * 13 / (20 * 19). So E[I_i] = 13 * 14 / (20 * 19).
By linearity of expectation,
E[∑119Ii]=∑119E[Ii]=20×1919×13×14=13×14/20=91/10