How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\](Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$.)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$
We can first consider the equation without a floor function:
\[\dfrac{n+1000}{70} = \sqrt{n}\]
Multiplying both sides by 70 and then squaring:
\[n^2 + 2000n + 1000000 = 4900n\]
Moving all terms to the left:
\[n^2 - 2900n + 1000000 = 0\]
Now we can determine the factors:
\[(n-400)(n-2500) = 0\]
This means that for $n = 400$ and $n = 2500$, the equation will hold without the floor function.
Now we can simply check the multiples of 70 around 400 and 2500 in the original equation:
For $n = 330$, left hand side  $=19$ but $18^2  23^2$ so right hand side  $=23$
Now we move to $n = 2500$
For $n = 2430$, left hand side  $=49$ and $49^2 < 2430 < 50^2$ so right hand side  $=49$
For $n = 2360$, left hand side  $=48$ and $48^2 < 2360 < 49^2$ so right hand side  $=48$
For $n = 2290$, left hand side  $=47$ and $47^2 < 2360 < 48^2$ so right hand side  $=47$
For $n = 2220$, left hand side  $=46$ but $47^2 < 2220$ so right hand side  $=47$
For $n = 2500$, left hand side  $=50$ and right hand side  $=50$
For $n = 2570$, left hand side  $=51$ but $2570 < 51^2$ so right hand side  $=50$
Therefore we have 6 total solutions, $n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{\textbf{(C) 6}}$