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 $x$.)

$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$
1. Let $\lfloor \sqrt{n} \rfloor = k$. This implies that $k \leq \sqrt{n} < k+1$, which translates to:
   \[
   k^2 \leq n < (k+1)^2
   \]
2. Given the equation:
   \[
   \frac{n + 1000}{70} = k
   \]
   we can solve for $n$:
   \[
   n = 70k - 1000
   \]
3. Substituting $n = 70k - 1000$ into the inequality $k^2 \leq n < (k+1)^2$, we get:
   \[
   k^2 \leq 70k - 1000 < (k+1)^2
   \]
4. We need to solve the inequalities separately:
   \[
   k^2 \leq 70k - 1000
   \]
   and
   \[
   70k - 1000 < (k+1)^2
   \]

5. For the first inequality:
   \[
   k^2 - 70k + 1000 \leq 0
   \]
   This quadratic equation can be factored as:
   \[
   (k - 20)(k - 50) \leq 0
   \]
   The solution to this inequality is:
   \[
   20 \leq k \leq 50
   \]

6. For the second inequality:
   \[
   70k - 1000 < k^2 + 2k + 1
   \]
   Simplifying, we get:
   \[
   k^2 - 68k + 1001 > 0
   \]
   Solving the quadratic equation $k^2 - 68k + 1001 = 0$ using the quadratic formula:
   \[
   k = \frac{68 \pm \sqrt{68^2 - 4 \cdot 1001}}{2} = \frac{68 \pm \sqrt{4624 - 4004}}{2} = \frac{68 \pm \sqrt{620}}{2} = \frac{68 \pm 2\sqrt{155}}{2} = 34 \pm \sqrt{155}
   \]
   Approximating $\sqrt{155} \approx 12.45$, we get:
   \[
   k \approx 34 \pm 12.45
   \]
   Thus, the roots are approximately:
   \[
   k \approx 21.55 \quad \text{and} \quad k \approx 46.45
   \]
   The solution to the inequality $k^2 - 68k + 1001 > 0$ is:
   \[
   k < 21.55 \quad \text{or} \quad k > 46.45
   \]

7. Combining the two solution sets:
   \[
   20 \leq k \leq 21.55 \quad \text{or} \quad 46.45 \leq k \leq 50
   \]
   The integer values of $k$ in these ranges are:
   \[
   k = 20, 21, 47, 48, 49, 50
   \]

8. For each value of $k$, we can find a corresponding $n$:
   \[
   n = 70k - 1000
   \]
   We need to verify that each $n$ lies within the interval $[k^2, (k+1)^2 - 1]$:
   - For $k = 20$: $n = 400$, and $20^2 \leq 400 < 21^2$
   - For $k = 21$: $n = 470$, and $21^2 \leq 470 < 22^2$
   - For $k = 47$: $n = 2290$, and $47^2 \leq 2290 < 48^2$
   - For $k = 48$: $n = 2360$, and $48^2 \leq 2360 < 49^2$
   - For $k = 49$: $n = 2430$, and $49^2 \leq 2430 < 50^2$
   - For $k = 50$: $n = 2500$, and $50^2 \leq 2500 < 51^2$

Thus, there are 6 values of $n$ that satisfy the given equation.

The final answer is $\boxed{6}$