Consider the sum:
\[
  S_n = \sum_{k = 1}^n \frac{1}{\sqrt{2k-1}} 
\]
Determine \( \lfloor S_{4901} \rfloor \). Recall that if \( x \) is a real number, then \( \lfloor x \rfloor \) (the floor of \( x \)) is the greatest integer that is less than or equal to \( x \).

Let's think step by step and output the final answer within \boxed{}.
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