Just take the conjugate!

Observe the following:

$\dfrac {1}{\sqrt{n}} = \dfrac{2}{\sqrt{n}+\sqrt{n}} < \dfrac {2}{\sqrt{n}+\sqrt{n-1}}$

Taking the conjugate, we obtain,

$\dfrac {1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$

Adding for $n=2, 3, ..., 10000$ , we get

$\dfrac {1}{\sqrt{2}} + \dfrac {1}{\sqrt{3}} + \ldots + \dfrac {1}{\sqrt{10000}} < 2(\sqrt{10000}-1)=198$

Similarly, observe

$\dfrac {1}{\sqrt{n}} = \dfrac{2}{\sqrt{n}+\sqrt{n}} > \dfrac {2}{\sqrt{n+1}+\sqrt{n}}$

Taking the conjugate, we obtain,

$\dfrac {1}{\sqrt{n}} > 2(\sqrt{n+1}-\sqrt{n})$

Adding for $n=2, 3, ..., 10000$ , we get

$\dfrac {1}{\sqrt{2}} + \dfrac {1}{\sqrt{3}} + \ldots + \dfrac {1}{\sqrt{10000}} > 2(\sqrt{10001}-\sqrt{2})>197$

Therefore,

$197 < \dfrac {1}{\sqrt{2}} + \dfrac {1}{\sqrt{3}} + \ldots + \dfrac {1}{\sqrt{10000}} < 198$

Thus, the value of

$\left \lfloor \dfrac {1}{\sqrt{2}} + \dfrac {1}{\sqrt{3}} + \ldots + \dfrac {1}{\sqrt{10000}} \right \rfloor = 197.$

Using upper and lower Riemann sums of the decreasing function $\frac{1}{\sqrt{x}}$ , we see that

$197<200-2\sqrt{2}=\int_{2}^{10000}\frac{dx}{\sqrt{x}}<\sum_{k=2}^{10000}\frac{1}{\sqrt{k}}<\int_{1}^{10000}\frac{dx}{\sqrt{x}}=198$

Thus the answer is $\boxed{197}$

This is how I thought of it (not a completely legit method but my general idea for the problem, and it worked here.)

If you take the average value of a set of numbers (let's say 5 numbers) and multiply the average by 5, you should still get the sum of those 5 numbers. So I was thinking, if you take consider this as $y = \dfrac{1}{\sqrt{x}}$ , I can use calculus to find the average value of that function from $[ 2, 10000 ]$ and multiply the average by $9999$ . So then, if I get the average value of $y$ , then that average value should be pretty close to the average value all terms in the given sum (but of course this needs to be proven more rigorously and I'm not going to do that right now. This could actually become another problem in itself with proving that the error is not more than ...)

$= \displaystyle \dfrac{1}{10000-2} \int_{2}^{10000} \dfrac{dx}{\sqrt{x}}$ $= \displaystyle \left. \dfrac{2 \sqrt{x}}{9998} \right|_{2}^{10000} = \dfrac{200-2\sqrt{2}}{9998}$

That gives about the average value of all the terms in the sum. Then, since there are $9999$ terms in the sum, multiply the average value by $9999$ and that gives about $197.2$ , and $\left \lfloor 197.2 \right \rfloor = \boxed{197}$ .