Incorrect cancellation of fraction with correct answer If you cancel (incorrectly) the 6 in $\frac{26}{65}$, you get the (correct) equation $\frac{26}{65} = \frac{2}{5}$. What other fraction exhibits this property?
I tried considering only the simple case where the fraction is of the form $\frac{10a+n}{10n+b}$ (as in the example), then attempting to solve $\frac{a}{b} = \frac{10a+n}{10n+b}$ to get $n(10a-b) = 9ab$. I was thinking of using divisibility tricks, but am not getting anywhere.
The case of the forms $\frac{10n+a}{10n+b}$ and $\frac{10a+n}{10b+n}$ are easy; you just get $a = b$. I haven't attempted higher than two digits or non-rationals. It would be interesting to see a complicated formula not necessarily involving a ratio of integers which works out to a correct answer with incorrect cancellation.

We'll solve $$\frac{10 a + n}{10 n + b} = \frac{a}{b},$$ or equivalently $n (10 a - b) = 9 a b$; to interpret the l.h.s. as a ratio of $2$-digit numbers, we restrict to solutions for which $1 \leq a, b, n \leq 9$.
Reducing modulo $9$ reduces $n(10 a - b) = 9 ab$ to $$n (a - b) \equiv 0 \pmod 9.$$ In particular $n$ and $(a - b)$ must have two factors of $3$ between them. This gives three cases:
CASE 1 $(3^2) \mid n$ : This forces $n = 9$, and canceling gives $10 a - b = ab$. Rearranging gives $(a + 1)(10 - b) = 10$, so since $1 \leq a, b \leq 10$ we must have either $a + 1 = 10$, or one of $a + 1$ and $10 - b$ is $2$ and the other is $5$. These cases respectively give $$\frac{99}{99}, \quad \frac{19}{95}, \quad \frac{49}{98}.$$
CASE 2 $3 \mid n$ and $3 \mid (a - b)$: We may as well assume $9 \nmid n$, as this is the content of case $1$, so, $n = 3$ or $n = 6$. If $n = 3$, simplifying gives $$10 a - b = 3 a b ,$$ and rearranging gives $(3 a + 1)(10 - 3 b) = 10$, which forces $a = b = 3$, giving $$\frac{33}{33} .$$ If $n = 6$, simplifying and rearranging gives $(3 a + 2)(20 - 3 b) = 40$. The factors of $40$ of the form $3 a + 2$ for $1 \leq a \leq 9$ are $5, 8, 20$, and these lead to $a = 1, b = 4$ and $a = 2, b = 5$, $a = 6, b = 6$, which respectively give
$$\frac{16}{64}, \quad \frac{26}{65}, \quad \frac{66}{66}.$$ 
CASE 3 $9 \mid a - b$: Since $1 \leq a, b \leq 9$, we have $a = b$, and so $n (10 a - b) = 9 a b$ simplifies to $n = a$, giving the trivial cases
$$\frac{11}{11}, \ldots, \frac{99}{99} .$$
In summary, the nontrivial solutions are
$$\color{#bf0000}{\boxed{\frac{16}{64}, \qquad \frac{19}{95}, \qquad \frac{26}{65}, \qquad \frac{49}{98}}}. $$
Of course, we can also look for "false cancellations" of the form $$\frac{10 n + a}{10 b + n} = \frac{a}{b} ,$$ but this is exactly the equation formed by taking the reciprocals of both sides of the above equation and interchanging the roles of $a, b$, so this equation leads precisely to the four reciprocals, $\frac{64}{16}$, etc., of the above solutions.
(There are other two-digit "false cancellation" fractions, corresponding to other equations, but there are all trivial in some sense; they have the forms $\frac{d0}{e0}$ and $\frac{dd}{ee}$ for digits $d, e$.)
