Four, (20 points) We know that $\frac{49}{98}$ simplifies to $\frac{1}{2}$, but by the method $\frac{49}{98}$, it also results in $\frac{1}{2}$. Find all proper fractions where both the numerator and the denominator are two-digit positive integers in decimal, the unit digit of the numerator is the same as the tens digit of the denominator, and they have the above "strange" property.
Let the true fraction $\frac{a b}{-}$ have the given property. Then $b c$
$\overline{a b}6$ when, $0<2 a-12<3 a+2$, at this time, $c$

is not an integer;
When $a=6$, $c=6$, contradiction;
When $a<6$, $12-2 a \geqslant 3 a+2$, so, $a \leqslant$ 2, and when $a=1$, $c=4$, at this time, the true fraction that satisfies the condition is $\frac{16}{64}$, when $a=2$, $c=5$, at this time, the true fraction that satisfies the condition is $\frac{26}{65}$.
(3) If $b=9$, then $a c=10 a-c$.
Thus, $c=\frac{10 a}{a+1}=10-\frac{10}{a+1}$.
Therefore, $(a+1) \mid 10$, so $a=1,4,9$.
When $a=1$, $c=5$, at this time, the true fraction that satisfies the condition is $\frac{19}{95}$;

When $a=4$, $c=8$, at this time, the true fraction that satisfies the condition is $\frac{49}{98}$;
When $a=9$, $c=9$, contradiction.
In summary, the true fractions that satisfy the condition are
$\frac{16}{64}, \frac{26}{65}, \frac{19}{95}, \frac{49}{98}$.