13. The answer is 12. The one-digit positive numbers with the reflective property are the perfect squares between 1 and 9, that is, 1, 4, and 9. There are no two-digit numbers with the reflective property: the perfect squares of two digits have a one-digit square root, which implies that they would enjoy the reflective property only if their two digits were the same; but a square with exactly two identical digits is divisible by 11, and thus by 121, and therefore does not exist. As for the three-digit numbers $ABC$ with the reflective property, they must be the squares of a certain two-digit number $XY$ and the number $100C + 10B + A$ must be the square of $10Y + X$. A two-digit number greater than 31 has a square greater than 1000, so both $10X + Y$ and $10Y + X$ must be at most 31. It follows that both $X$ and $Y$ must be at most 3, and if one of the digits is 3, the other must be 0 or 1. The only possibilities to consider for the square root of $ABC$ are therefore $10, 11, 12, 13, 20, 21, 22, 30, 31$, which produce the following squares with the reflective property: $100, 121, 144, 169, 400, 441, 484, 900, 961$. We thus have a total of 12 numbers of at most 3 digits with the reflective property.
13. The number 144 is very special in that it enjoys the mirror property: not only is it true that $144=12^{2}$ (i.e., it is a perfect square), but it is also true that $441=21^{2}$, meaning that reversing the order of its digits results in precisely the square of the number obtained by reversing the digits of its square root. The number 100 also has the mirror property, as does the number 1, while the number 49 does not. How many positive integers with at most 3 digits enjoy the mirror property?