If $N$ has one digit this is not possible. If $N$ has two digits, say $N = ab$, then $\overline\{N\} = ba$ so that $N - \overline\{N\} = 9(a - b)$ which yields $a - b = 1$ so the numbers are
$$N = 10, 21, 32, 43, 54, 65, 76, 87, 98.$$
Assume there are numbers $N$ with the above property that have $m \ge 3$ digits. Thus, we may write
$$N = aZb$$
where $a \in \{1, 2, 3, \dots, 9\}$ and $b \in \{0, 1, 2, 3, \dots, 9\}$ are digits and $Z$ is a positive integer with $m - 2 \ge 1$ digits. We may need to include leading zeros in the presentation of $Z$. Then, $\overline\{N\} = b\overline\{Z\}a$, where $\overline\{Z\}$ is obtained by writing the digits of $Z$ from right to left. We now have
$$9 = N - \overline\{N\} = (10^{m-1}a + 10Z + b) - (10^{m-1}b + 10\overline\{Z\} + a)$$
which yields
$$9 = (10^{m-1} - 1)(a - b) + 10(Z - \overline\{Z\}). \quad (6)$$
Because $N \ge \overline\{N\}$, we must have $a \ge b$.
If $a - b = 0$ then (6) cannot hold because the right-hand side is a multiple of 10 while the left-hand side is not. If $a - b \ge 2$ we again raise a contradiction since $N \ge a \cdot 10^{m-1}$ and $\overline\{N\} \le (b + 1) \cdot 10^{m-1}$ imply
$$9 = N - \overline\{N\} \ge (a - b - 1) \cdot 10^{m-1} \ge 10^{m-1} \ge 100.$$
Hence $a - b = 1$ and from (6) we find $9 = 10^{m-1} - 1 + 10(Z - \overline\{Z\})$ from which we get
$$\overline\{Z\} - Z = 10^{m-2} - 1. \quad (7)$$
Clearly, $Z = 0$ does not satisfy (7). If $Z > 0$, (7) implies $\overline\{Z\} > 10^{m-2} - 1$. Recall now that $Z$ and $\overline\{Z\}$ have at most $m - 2$ digits, fewer than $m - 2$ if $Z$ contains leading digits equal to zero. Hence, $Z$ and $\overline\{Z\}$ cannot exceed $10^{m-2} - 1 = 999\dots9$, the number that consists of $m - 2$ digits 9. This contradiction shows that when $m \ge 3$ there is no solution.
Hence, the only numbers satisfying the condition in the problem are the two-digit numbers listed above.