The problem is the following. Let's say we have two boxes, each with $R$ red balls and $B$ blue balls, i.e $R+B$ balls in each of them. We draw pairs of balls, choosing one from each of the boxes until there are no more balls left. Given $X$ the total number of red and blue balls pairs drawn, I am trying to find $E(X)$.
I considered this approach. We start by defining new variables $X_1,...,X_{R+B}$ where $X_i = 1$ if the pair of balls drawn in the ith extraction is blue and red (regardless of their order) and $0$ if it isn't. Now by linearity of expectation we have that: $$E(X) = \sum_{i=1}^{R+B} E(X_i) = \sum_{i=1}^{R+B}P(X_i = 1)$$
Now $P(X_1 = 1) = \frac{2RB}{(R+B)^2}$. My problem is that finding $P(X_i = 1)$ for $i > 1$ seems to be much harder since I think that I would have to condition the probability over all the possible previous extractions. I was wondering whether that is the best approach or there is an easier way to find $E(X)$.
Linearity suffices. The beauty of linearity is that it holds even if, as here, the variables are dependent.
Any given pair has probability $\frac {2RB}{(R+B)^2}$ as you remark, so the answer is simply $$E[X]=(R+B)\times \frac {2RB}{(R+B)^2}=\frac {2RB}{(R+B)}$$
As a quick sanity check, suppose $(R,B)=(1,1)$. Then, if the first pair is mixed (a $\frac 12$ chance), the second is also. And if the first is single colored, so is the second. Thus you have two mixed pairs or none with equal chances, so the expected number is $1$, as the formula implies.