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 [i]floor[/i] of $x$) is the greatest integer that is less than or equal to $x$.

1. **Estimate the sum using an integral approximation:**

   We start by approximating the sum \( S_n = \sum_{k=1}^n \frac{1}{\sqrt{2k-1}} \) using an integral. The idea is to approximate the sum by the integral of a similar function. Specifically, we use:
   \[
   \sum_{k=1}^n \frac{1}{\sqrt{2k-1}} \approx \int_1^n \frac{dx}{\sqrt{2x-1}}
   \]

2. **Simplify the integral:**

   To simplify the integral, we make a substitution. Let \( u = 2x-1 \), then \( du = 2dx \) or \( dx = \frac{du}{2} \). The limits of integration change accordingly: when \( x = 1 \), \( u = 1 \); and when \( x = n \), \( u = 2n-1 \). Thus, the integral becomes:
   \[
   \int_1^n \frac{dx}{\sqrt{2x-1}} = \int_1^{2n-1} \frac{du}{2\sqrt{u}} = \frac{1}{2} \int_1^{2n-1} u^{-\frac{1}{2}} \, du
   \]

3. **Evaluate the integral:**

   The integral \( \int u^{-\frac{1}{2}} \, du \) is straightforward:
   \[
   \int u^{-\frac{1}{2}} \, du = 2u^{\frac{1}{2}} = 2\sqrt{u}
   \]
   Applying the limits of integration, we get:
   \[
   \frac{1}{2} \left[ 2\sqrt{u} \right]_1^{2n-1} = \left[ \sqrt{u} \right]_1^{2n-1} = \sqrt{2n-1} - \sqrt{1} = \sqrt{2n-1} - 1
   \]

4. **Refine the approximation:**

   The initial approximation can be refined by considering the excess on the interval \([0,1]\). A better estimate is:
   \[
   S_n \approx 1 + \int_1^n \frac{dx}{\sqrt{2x-1}}
   \]
   Using the previous result, we have:
   \[
   S_n \approx 1 + (\sqrt{2n-1} - 1) = \sqrt{2n-1}
   \]

5. **Apply the approximation to \( S_{4901} \):**

   For \( n = 4901 \):
   \[
   S_{4901} \approx \sqrt{2 \cdot 4901 - 1} = \sqrt{9801} = 99
   \]

6. **Estimate the error:**

   To rigorize the approximation, we estimate the error. The error in the interval \([m, m+1]\) for \( m = 1, 2, \ldots, n-1 \) is:
   \[
   \int_m^{m+1} \frac{dx}{\sqrt{2x-1}} - \frac{1}{\sqrt{2m-1}}
   \]
   Using the integral approximation, the error is:
   \[
   \frac{\sqrt{2}}{\sqrt{m+1} + \sqrt{m}} - \frac{1}{\sqrt{2m-1}}
   \]
   Simplifying, we get:
   \[
   \frac{\sqrt{4m+2} - \sqrt{m+1} - \sqrt{m}}{\sqrt{2m-1}(\sqrt{m+1} + \sqrt{m})} < \frac{\sqrt{6} - \sqrt{2} - 1}{2\sqrt{2}m}
   \]

7. **Sum the errors:**

   The total error is at most:
   \[
   \sum_{m=1}^{n-1} \frac{\sqrt{6} - \sqrt{2} - 1}{2\sqrt{2}m} = \left( \frac{\sqrt{3}}{2} - \frac{1}{2} - \frac{\sqrt{2}}{4} \right) H_{n-1}
   \]
   where \( H_{n-1} \) is the \((n-1)\)-th harmonic number. Using the approximation \( H_{n-1} \approx \log(n-1) + \gamma \) (where \(\gamma\) is the Euler-Mascheroni constant), we get:
   \[
   \left( \frac{\sqrt{3}}{2} - \frac{1}{2} - \frac{\sqrt{2}}{4} \right) (\log(n-1) + 1) < 0.13
   \]

8. **Final bounds:**

   Thus, the value of the sum is between:
   \[
   \sqrt{9802} - \sqrt{2} + 1 - 0.13 > 98 \quad \text{and} \quad \sqrt{9802} - \sqrt{2} + 1 < 99
   \]
   Therefore, the floor of the sum is:
   \[
   \lfloor S_{4901} \rfloor = 98
   \]

The final answer is \(\boxed{98}\)