If $n=81$, then $\sqrt n=9$, so numbers close to $81$ will work.

What other square numbers are close to $81$? We know that $\sqrt{64}=8$, so $n$ can't be $64$. But if $n$ is a little bit larger than $64$, then $\sqrt n$ will be just larger than $8$, so the distance between $\sqrt n$ and $9$ will be less than $1$.

We also know that $\sqrt{100}=10$, so $n$ can't be $100$. But if $n$ is a little bit smaller than $100$, then $\sqrt n$ will be just smaller than $10$, so the distance between $\sqrt n$ and $9$ will be less than $1$.

So $n$ can be any number from $65$ to $99$, which is a total of $35$ numbers.

If $\sqrt n < 9$, then we must have $9-\sqrt n<1$.

If $\sqrt n > 9$, then we must have $\sqrt n-9<1$.

Solving these two inequalities will give us the possible values of $n$.

$$\begin{align} 9-\sqrt n &<1\\

\Rightarrow 9&<1+\sqrt n\\

\Rightarrow 8&<\sqrt n\end{align}$$ Because $8$ and $\sqrt n$ are both positive, we can square both sides without risking multiplying by a negative number (which would change the inequality sign!). So $$8^2<\sqrt n^2\Rightarrow 64<n$$

Similarly, we can solve the other inequality:$$ \sqrt n -9 <1\Rightarrow \sqrt n<10$$ Again, since $10$ and $\sqrt n$ are both positive, we can square both sides without risking multiplying by a negative number: $$\sqrt n^2<10^2\Rightarrow n<100$$

So $n$ is greater than $64$ but less than $100$, which means that $n$ can be any number from $65$ to $99$. That is a total of $35$ numbers.

You can find more short problems, arranged by curriculum topic, in our short problems collection.