IRL_ABooklet

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 $$9=(10^{m-1}-1)(a-b)+10(Z-\overline{\{}Z\}).(6)$$ 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 $$\overline{\{}Z\}-Z=10^{m-2}-1.(7)$$ 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 $10^{m-2}-1=999\dots 9$, 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.