The tens digit is quite odd...

Let us represent $n$ as $100a+b$ . Squaring this, we get $10000a^2+200ab+b^2$ . Since $10000a^2$ and $200ab$ do not affect the tens digit, we only need to concern about $b^2$ . Again, we can represent $b$ as $10x+y$ . Proceeding, $b^2=100x^2+20xy+y^2$ . The $100x^2$ does not affect the tens digit. The $20xy$ has an even tens digit, so in order for the tens digit of $n^2$ to be odd, the tens digit of $y^2$ must also be odd. Since $y$ is a single-digit positive integer, we only need to consider the tens digit of the squares of the integers from $1$ to $9$ :

$4$ and $6$ are the only ones that have squares with an odd tens digit ( $1$ and $3$ , respectively). Therefore, all numbers with units digit of either $4$ or $6$ will have an odd tens digit. Simple counting gives us $\boxed{200}$ total values of $n$ .