Inspired by Comrade Quan

Define the function $f(x)=\frac{10^4}{\sqrt{x}}$ and consider the trapezoidal sum of $f(x)$ between $a=10^4$ and $b=10^6$ with step size 1. Since the graph of $f(x)$ is convex, this sum will exceed the integral: $S-\frac{f(10^4)}{2}-\frac{f(10^6)}{2}=S-50-5>10^4\int_{10^4}^{10^6}x^{-1/2}dx=1.8\times10^7$ so $S>18000055.$

The maximum of $f''(x)$ on $[a,b]$ is $M=7.5\times 10^{-7}$ , so that the error is $<\frac{(b-a)M}{12}<\frac{1}{10}$ . Thus the floor of $S$ is $\boxed{18000055}$ ... an estimate with remarkable accuracy.