Then answer is $\boxed{1998}$. It is easy to verify that$$\frac{2}{\sqrt{n} + \sqrt{n-1}} > \frac{1}{\sqrt{n}} > \frac{2}{\sqrt{n+1} + \sqrt{n}}$$ $$\implies 2(\sqrt{n} - \sqrt{n-1}) > \frac{1}{\sqrt{n}} > 2(\sqrt{n+1} -\sqrt{n})$$ Therefore $$ \begin{array}{rcccl} 1 &=& 1 &=& 1\\ {2}(\sqrt{2}-\sqrt{1}) & > & \frac{1}{\sqrt{2}} & > & {2}(\sqrt{3}-\sqrt{2})\\ {2}(\sqrt{3}-\sqrt{2}) & > & \frac{1}{\sqrt{2}} & > & {2}(\sqrt{4}-\sqrt{3})\\ &&\cdots\\ 2(\sqrt{1000000} -\sqrt{999999} & > & \frac{1}{\sqrt{1000000}} & > & 2(\sqrt{1000001} - \sqrt{1000000}) \end{array} $$ Adding these together we find $$2\sqrt{1000000} -1 > 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{1000000}} > 2\sqrt{10000001} - 2\sqrt{2}+1$$ $$\implies 199 > 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{1000000}} > 1998.173...$$