So, the given distribution is a discrete $(X,Y)$ vector with the following probability mass function.
$$ \mathbb{P}(X=x, Y=y)=\frac{1}{2 n+1} \frac{\left(\begin{array}{c} n \\ x \end{array}\right)\left(\begin{array}{l} n \\ y \end{array}\right)}{\left(\begin{array}{c} 2 n \\ x+y \end{array}\right)} \mathbb{I}_{\{0, \ldots, n\}}(x) \mathbb{I}_{\{0, \ldots, n\}}(y) $$
For a given $n$. It looks like a joint hypergeometric my intuition says. My approach was to calculating the distribution was:
$$ \begin{aligned} P(X=x \mid X+Y=t)&=P(X=x \mid Y=t-x) \\ & =\frac{P(X=x, Y=t-x)}{P(Y=t-x)} \end{aligned} $$
And, with those requirements, I just need to calculate $P(Y=y)$ by marginalizing it with respect to $X$:
$$ \begin{aligned} & P(Y=y)=\sum_{x=0}^n P(x=x, y=y)= \\ & =\sum_{x=0}^n \frac{1}{2 n+1} \frac{\left(\begin{array}{l} n \\ x \end{array}\right)\left(\begin{array}{l} n \\ y \end{array}\right)}{\left(\begin{array}{c} 2 n \\ x+y \end{array}\right)} \cdot I_{\{1, \cdots ,n\}}(y) \\ & =\frac{1}{2 n+1}\left(\begin{array}{l} n \\ y \end{array}\right) \cdot I_{\{1, \cdots, n\}}(y) \sum_{x=0}^n \frac{\left(\begin{array}{l} n \\ x \end{array}\right)}{\left(\begin{array}{l} 2 n \\ (x+y) \end{array}\right)} \end{aligned} $$ But I don't know how to handle the summation. Is there a property for binomial coefficients I'm not aware, or a different approach to reaching the distribution? In the original question, they mention the formula for the expectation of a hypergeometric, hinting that that is what the conditional distribution boils down to.