Find the number of positive integers n such that $\sqrt{n+\sqrt{n+\sqrt{n}}}<10$ for any finite number of square root signs. Find the number of positive integers $n$ such that $\sqrt{n+\sqrt{n+\sqrt{n}}}<10$ for any finite number of square root signs.
I know something with squaring and repeating, but what does "finite number of square root signs" mean?

It seems that the question means to ask you, as mentioned by the user lulu, that: 
Find all natural numbers $n$ such that each of $\sqrt n,\sqrt{n+\sqrt n}, \sqrt{n+\sqrt{n+\sqrt n}},\dots$ is less than 10. 
Which is a round-about way of asking:
Find all natural numbers $n$ such that $\sqrt{n+\sqrt{n+\sqrt {n+\dots}}}<10$.
To solve it, we may use the standard operating procedure:
$$\text{Let }\sqrt{n+\sqrt{n+\sqrt {n+\dots}}}=x$$
$$\Rightarrow\sqrt{n+x}=x$$
$$\Rightarrow x^2-x-n=0$$
$$\Rightarrow x=\frac{1\pm\sqrt{1+4n}}{2}$$
And we are given that $x<10$. So:
$$\frac{1\pm\sqrt{1+4n}}{2}<10$$
$$\pm\sqrt{1+4n}<19$$
$$1+4n<361$$
$$n<90$$
But for an infinite sequence would $n=90$ not be less than $10$. Thus, for a finite sequence $n=90$ shouldn't be a problem.
Had we got something like $n<89.92$ then we would have concluded that max value $n$ can assume is $89$.
Thus, for your case, $n$ can be any value from $1$ to $90$.
