jmc

If there are exactly $2$ positive two-digit multiples of $x$, those two multiples must be $x$ and $2x$. Therefore, $2x$ must be less than $100$, while $3x$ the next largest multiple of $x$, must be at least $100$ (or else there would be $3$, not $2$ multiples in the two-digit range).

It may take some trial and error to find the smallest and largest possible values of $x$ under these conditions. The smallest is $x=34$, because $3x=102$, the smallest three-digit multiple of $3$. If we tried anything smaller than $34$, $x$, $2x$, and $3x$ would all have two digits, and that doesn't satisfy the condition.

The largest possible value of $x$ is $49$, because if $x$ were $50$, $2x$ would equal $100$, and only one multiple of $x$ would have two digits. Every value of $x$ from $34$ to $49$ works.

Now, we must count the number of integers from $34$ to $49,$ inclusive. This is a surprisingly tricky process: you might think there should be $49-34$, or $15$ possible values of $x$, but that's not actually right! Suppose we subtract $33$ from each number. Then we are counting the numbers from $1$ to $16,$ and so there are $\boxed{16}$ integers from $34$ to $49,$ inclusive.