Let $S = 1 + \frac{1}{\sqrt2}+ \frac{1}{\sqrt3}+\frac{1}{\sqrt4}+...+ \frac{1}{\sqrt{99}}+ \frac{1}{\sqrt{100}}$ . Find $[S]$.

You may use the fact that $\sqrt{n} < \frac12 (\sqrt{n} +\sqrt{n+1}) <\sqrt{n+1}$ for all integers $n \ge 1$.
1. We start with the given series \( S = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{100}} \).

2. We use the inequality \( \sqrt{n} < \frac{1}{2} (\sqrt{n} + \sqrt{n+1}) < \sqrt{n+1} \) for all integers \( n \geq 1 \). This can be rewritten as:
   \[
   \sqrt{n-1} < \frac{1}{2} (\sqrt{n} + \sqrt{n-1}) < \sqrt{n}
   \]

3. Taking the reciprocal of the terms, we get:
   \[
   \frac{1}{\sqrt{n}} < \frac{2}{\sqrt{n} + \sqrt{n-1}} < \frac{1}{\sqrt{n-1}}
   \]

4. Multiplying through by 2, we obtain:
   \[
   \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1}) < \frac{1}{\sqrt{n-1}}
   \]

5. Summing these inequalities from \( n = 2 \) to \( n = 100 \), we get:
   \[
   \sum_{n=2}^{100} \frac{1}{\sqrt{n}} > 2 \sum_{n=2}^{100} (\sqrt{n} - \sqrt{n-1})
   \]
   and
   \[
   \sum_{n=2}^{100} \frac{1}{\sqrt{n}} < \sum_{n=2}^{100} \frac{1}{\sqrt{n-1}}
   \]

6. The right-hand side of the first inequality simplifies as a telescoping series:
   \[
   2 (\sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2} + \cdots + \sqrt{100} - \sqrt{99}) = 2 (\sqrt{100} - \sqrt{1}) = 2 (10 - 1) = 18
   \]

7. Therefore, we have:
   \[
   \sum_{n=2}^{100} \frac{1}{\sqrt{n}} > 18
   \]

8. For the left-hand side of the second inequality, we note:
   \[
   \sum_{n=2}^{100} \frac{1}{\sqrt{n-1}} = \sum_{n=1}^{99} \frac{1}{\sqrt{n}}
   \]

9. Adding the first term \( 1 \) to both sides, we get:
   \[
   S = 1 + \sum_{n=2}^{100} \frac{1}{\sqrt{n}} > 1 + 18 = 19
   \]

10. Combining the inequalities, we have:
    \[
    18 < S < 19
    \]

11. Therefore, the integer part of \( S \) is:
    \[
    \lfloor S \rfloor = 18
    \]

The final answer is \(\boxed{18}\)