smc

We can first consider the equation without a floor function: $$\frac{n+1000}{70}=\sqrt{n}$$ 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, which we abbreviate as $L=R$. For $n=330$, $L=19$ but $18^2<330<19^2$ so $R=18$ For $n=400$, $L=20$ and $R=20$ For $n=470$, $L=21$, $R=21$ For $n=540$, $L=22$ but $540>23^2$ so $R=23$ Now we move to $n=2500$ For $n=2430$, $L=49$ and $49^2<2430<50^2$ so $R=49$ For $n=2360$, $L=48$ and $48^2<2360<49^2$ so $R=48$ For $n=2290$, $L=47$ and $47^2<2360<48^2$ so $R=47$ For $n=2220$, $L=46$ but $47^2<2220$ so $R=47$ For $n=2500$, $L=50$ and $R=50$ For $n=2570$, $L=51$ but $2570<51^2$ so $R=50$ Therefore we have 6 total solutions, $n=400,470,2290,2360,2430,2500=\boxed{6}$