To solve the series:

\[
S = \frac{5}{2 \times 4} + \frac{5}{4 \times 6} + \frac{5}{6 \times 8} + \cdots + \frac{5}{2024 \times 2025}
\]

we observe that it is a telescoping series, which can be simplified using partial fraction decomposition.

Each term in the series can be decomposed as follows:

\[
\frac{5}{n(n+2)} = \frac{A}{n} + \frac{B}{n+2}
\]

Solving for \(A\) and \(B\):

\[
5 = A(n+2) + Bn
\]

\(\Rightarrow 5 = An + 2A + Bn\)

\(\Rightarrow 5 = (A + B)n + 2A\)

Equating coefficients, we get:

1. \(A + B = 0\)
2. \(2A = 5\)

From equation 2, \(A = \frac{5}{2}\).

Substitute \(A\) into equation 1:

\(\frac{5}{2} + B = 0\)

\(\Rightarrow B = -\frac{5}{2}\)

Thus, the partial fraction decomposition is:

\[
\frac{5}{n(n+2)} = \frac{5/2}{n} - \frac{5/2}{n+2}
\]

So, each term in the series becomes:

\[
\frac{5}{n(n+2)} = \frac{5/2}{n} - \frac{5/2}{n+2}
\]

Substitute back into the series:

\[
S = \left(\frac{5/2}{2} - \frac{5/2}{4}\right) + \left(\frac{5/2}{4} - \frac{5/2}{6}\right) + \left(\frac{5/2}{6} - \frac{5/2}{8}\right) + \ldots + \left(\frac{5/2}{2024} - \frac{5/2}{2026}\right)
\]

This series is telescoping, meaning most terms cancel out:

\[
S = \frac{5/2}{2} - \frac{5/2}{2026}
\]

Simplifying:

\[
S = \frac{5}{4} - \frac{5}{2 \times 2026}
\]

\[
S = \frac{5}{4} - \frac{5}{4052}
\]

To simplify, find a common denominator (4052):

\[
S = \frac{5 \times 1013}{4052} - \frac{5}{4052}
\]

\[
S = \frac{5 \times 1012}{4052}
\]

We can reduce it:

Since \(1012\) and \(4052\) have no common factors other than 1, the final answer is:

\[
S = \frac{1265}{1013}
\]

Thus, the sum of the series is \(\frac{1265}{1013}\).