It is easy to verify that$$\frac{2}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{\sqrt{n}} <\frac{2}{\sqrt{n}+\sqrt{n-1}} $$ Simplifying the left and the right sides gives: $$2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$$ Hence $$\begin{array}{rccl} 1&=& 1 &=&1\\ 2(\sqrt{3} -\sqrt{2}) &<& \frac{1}{\sqrt{2}} &<& 2(\sqrt{2} - \sqrt{1})\\ 2(\sqrt{4} -\sqrt{3}) &<& \frac{1}{\sqrt{3}}&<&2(\sqrt{3} - \sqrt{2})\\ & \cdots &\\ 2(\sqrt{n+1} -\sqrt{n})& <& \frac{1}{\sqrt{n}} &<& 2(\sqrt{n} - \sqrt{n-1}) \end{array}$$ Adding these relations together gives $$2\sqrt{n+1}-2\sqrt{2}+1< 1 + \frac{1}{\sqrt{2}}+ \frac{1}{\sqrt{3}} + \cdots + + \frac{1}{\sqrt{n}} < 2\sqrt{n}-2+1$$ Setting $n=10000$ gives $$\begin{array}{rcccl} 2\sqrt{10001} -2\sqrt{2}+1 &<& 1 + \frac{1}{\sqrt{2}}+ \frac{1}{\sqrt{3}} + \cdots + + \frac{1}{\sqrt{10000}} &<& 2\sqrt{10000}-1\\ 198.1... &<& 1 + \frac{1}{\sqrt{2}}+ \frac{1}{\sqrt{3}} + \cdots + + \frac{1}{\sqrt{10000}} &<& 199 \end{array}$$ Therefore the answer is $\boxed{198}$.