The Arithmetic Mean-Geometric Mean-Harmonic Mean inequality states:
\[ \text{AM} \ge \text{GM} \ge \text{HM} \] which for two positive numbers \( x \) and \( y \) is:
\[ \frac{x + y}{2} \ge \sqrt{xy} \ge \frac{2}{\frac{1}{x} + \frac{1}{y}} \]
We are given that \( x + y = 102 \). To maximize the product \( xy \), we apply the AM-GM inequality:
\[ xy \le \left(\frac{x + y}{2}\right)^2 \] Substitute the given sum:
\[ xy \le \left(\frac{102}{2}\right)^2 = 51^2 = 2601 \] This implies an upper bound for \( xy \). Consequently, the reciprocal \( \frac{1}{xy} \) has a lower bound:
\[ \Rightarrow \frac{1}{xy} \ge \frac{1}{2601} \] From the HM-GM inequality, we have \( \sqrt{xy} \ge \frac{2}{\frac{1}{x} + \frac{1}{y}} \), which rearranges to \( \frac{1}{x} + \frac{1}{y} \ge \frac{2}{\sqrt{xy}} \). Since \( \sqrt{xy} \le 51 \), it follows that:
\[ \Rightarrow \frac{1}{x} + \frac{1}{y} \ge \frac{2}{51} \]
The expression to minimize is:
\[ 2601 \left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right) \] Expanding this expression yields:
\[ = 2601 \left(1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{xy}\right) \] To find the minimum value, we substitute the minimum possible values for the terms \( \frac{1}{x} + \frac{1}{y} \) and \( \frac{1}{xy} \) derived from the inequalities:
\[ \frac{1}{x} + \frac{1}{y} \ge \frac{2}{51}, \quad \frac{1}{xy} \ge \frac{1}{2601} \] Substituting these minimum values into the expanded expression:
\[ = 2601 \left(1 + \frac{2}{51} + \frac{1}{2601}\right) \]
Now, we simplify the resulting expression:
\[ = 2601 \left(1 + \frac{2}{51} + \frac{1}{2601} \right) \] Combine the terms inside the parenthesis by finding a common denominator:
\[ = 2601 \left( \frac{2601}{2601} + \frac{2 \times 51}{51 \times 51} + \frac{1}{2601} \right) = 2601 \left( \frac{2601}{2601} + \frac{102}{2601} + \frac{1}{2601} \right) \] \[ = 2601 \left(\frac{2601 + 102 + 1}{2601} \right) = 2601 \left(\frac{2704}{2601} \right) \] Cancel out the \( 2601 \):
\[ = 2704 \]
The minimum value of the expression is:
\[ \boxed{2704} \]