The secret code

Using $N\!=\!1,019,090$ , let's write $\displaystyle \sum_{n=1}^N\dfrac{1}{\sqrt{n}}\,=\,\dfrac{1}{\sqrt{1}}+\displaystyle \sum_{n=2}^N\dfrac{1}{\sqrt{n}}\,=\,1+\displaystyle \sum_{n=2}^N\dfrac{1}{\sqrt{n}}$ .

Because the function $f(x)\!=\!\dfrac{1}{\sqrt{n}}$ is decreasing, we have $\boxed{1+\displaystyle \int_2^{N+1}\!\!\dfrac{1}{\sqrt{x}}\,\mathrm{d}x\,\,\,<\,\,\,1+\displaystyle \sum_{n=2}^N\dfrac{1}{\sqrt{n}}\,\,\,<\,\,\,1+ \displaystyle \int_1^{N}\!\!\dfrac{1}{\sqrt{x}}\,\mathrm{d}x}$

$\displaystyle \int\!\!\dfrac{1}{\sqrt{x}}\,\mathrm{d}x\,=\,\int\!\!x^{-\frac12}\,\mathrm{d}x\,=\,\dfrac{x^\frac12}{^1\!/_2}+c=2\sqrt{x}+c$ .

We'll have to get the integer part of $2\sqrt{N}\text{ and }2\sqrt{N+1}$ , that is the square roots with a 0.5 precision.

Without any calculator, we can see that $(1,000+x)^2=1,000,000+2,000x+x^2$ , so $x$ must be around $\frac{19,090}{2,000}\simeq9.5$ .

$1,009^2=1,000,000+18,000+81<1,019,090$ .

Let's see what gives $1009.5^2=(1009+0.5)\cdot(1010-0.5)=1009\cdot1010+0.5\cdot(1010-1009)-0.5^2=1009\cdot(1000+10)+0.5\cdot1-0.25=1,009,000+10,090+0.25=1,019,090.25$ .

Therefore, $\sqrt{N}$ is just a bit short of $1009.5$ , that is $\sqrt{N}=1009.49\!\ldots\implies2\sqrt{N}=2018.9\!\ldots$

and $\sqrt{N+1}$ is just after $1009.5$ , that is $\sqrt{N+1}=1009.5\!…\implies2\sqrt{N+1}=2019.0\!…$

$\displaystyle \int_2^{N+1}\!\!\dfrac{1}{\sqrt{x}}\,\mathrm{d}x\,=\,\Big[2\sqrt{x}\,\Big]_2^{N+1}\!=\,2\sqrt{N+1}-2\sqrt2\,=\,2019.0\!…-2.828\!…>2016$

$\displaystyle \int_1^{N}\!\!\dfrac{1}{\sqrt{x}}\,\mathrm{d}x\,=\,\Big[2\sqrt{x}\,\Big]_1^{N}\!=\,2\sqrt{N}-2\sqrt1\,=\,2018.9\!…-2<2017$ .

Therefore, $\underbrace{1+2016}_{2017}\,<\,1+\displaystyle \sum_{n=2}^N\dfrac{1}{\sqrt{n}}\,<\,\underbrace{1+ 2017}_{2018}$ , which gives $\displaystyle \sum_{n=1}^N\dfrac{1}{\sqrt{n}}=2017.\,…$ , that is, rounded down, $\boxed{\text{answer}=2017}$ .