We are given that:
\[
a_n = \frac{n^2 + 5n + 6}{4}
\]
Now, we calculate \( S_n \) as:
\[
S_n = \sum_{k=1}^{n} \frac{1}{a_k} = \sum_{k=1}^{n} \frac{4}{k^2 + 5k + 6}
\]
We can break this sum down into partial fractions:
\[
S_n = 4 \sum_{k=1}^{n} \frac{1}{(k+2)(k+3)}
\]
This is equivalent to:
\[
S_n = 4 \sum_{k=1}^{n} \left( \frac{1}{k+2} - \frac{1}{k+3} \right)
\]
Thus, we have:
\[
S_n = 4 \left( \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \ldots \right)
\]
The sum for \( S_n \) will cancel out and simplify as follows:
\[
S_{2025} = 4 \left( \frac{1}{3} - \frac{1}{2028} \right)
\]
Now calculate the value of \( 507 S_{2025} \):
\[
507 S_{2025} = 507 \times 4 \times \left( \frac{1}{3} - \frac{1}{2028} \right) = 675
\]