Given that, $x$ and $y$ are positive real numbers, satisfying $x+y = 102.$
We know that, $\boxed{\text{AM} \geq \text{GM} \geq \text{HM}}$
$ \Rightarrow \frac{x+y}{2} \geq \sqrt{xy} \geq \dfrac{2}{\frac{1}{x} + \frac{1}{y}} $
$ \Rightarrow \frac{102}{2} \geq \sqrt{xy} \geq \dfrac{2}{\frac{1}{x} + \frac{1}{y}} \quad \longrightarrow (1)$
Take the first two terms, from equation $(1).$
$ 51 \geq \sqrt{xy} $
$ \Rightarrow xy \leq 51^{2} $
$ \Rightarrow \boxed{xy \leq 2601} $
Taking the last two terms, from equation $(1).$
$ \sqrt{xy} \geq \dfrac{2}{\frac{1}{x} + \frac{1}{y}} $
$ \Rightarrow 51 \geq \dfrac{2}{\frac{1}{x} + \frac{1}{y}} $
$ \Rightarrow \boxed{\frac{1}{x} + \frac{1}{y} \geq \frac{2}{51}} $
Now, the minimum possible value of $ 2601 \left( 1 + \frac{1}{x} \right) \left( 1 + \frac{1}{y} \right) = 2601 \left( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy} \right) $
$\qquad = 2601 \left( 1 + \frac{2}{51} + \frac{1}{2601} \right) = 2601 \left( \frac{2601 + 102 + 1}{2601} \right) = 2704 $
$\therefore$ The minimum possible value is $2704.$
Correct Answer$: 2704$