jmc

First notice that the graphs of $(n+1000)/70$ and $\sqrt[]{n}$ intersect at 2 points. Then, notice that $(n+1000)/70$ must be an integer. This means that n is congruent to $50(\mathrm{mod}70)$. For the first intersection, testing the first few values of $n$ (adding $70$ to $n$ each time and noticing the left side increases by $1$ each time) yields $n=20$ and $n=21$. Estimating from the graph can narrow down the other cases, being $n=47$, $n=50$. This results in a total of $\boxed{6}$ cases.