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Answer: $39$

$\sqrt {n}$ needs to be between $9$ and $11$
$\therefore n$ needs to be between $81$ and $121$
So $n= 82, 83, ... , 119, 120$

$\overbrace{\underbrace{1, 2, 3, ..., 80, 81}_{81\text{ numbers}}, 82, ..., 120}^{120\text{ numbers}}$

So $82, ..., 120$ is $120-81=39$ numbers.


 

This problem is taken from the UKMT Mathematical Challenges.
You can find more short problems, arranged by curriculum topic, in our short problems collection.