The sum and difference of 44 and 18 consist of the same digits, just in reverse order. How many pairs of two-digit numbers have this property?
Let the sought numbers be denoted by $A$ and $B$, their sum by $S$, and their difference by $F$. Since the digits of $S$ and $F$ are the same, only in reverse order, both are two-digit numbers, as $S$ is at least two digits, and $F$ is at most two digits. Since $F>0$, $A$ is greater than $B$, and if the digits of $S$ are denoted by $x$ and $y$, $x$ is greater than $y$. Taking into account their place values, the values of $S$, $F$, $A$, and $B$ are as follows:

$$
\begin{aligned}
& S=10 x+y, \quad F=10 y+x \\
& A=\frac{1}{2}(S+F)=11 \frac{x+y}{2} \\
& B=\frac{1}{2}(S-F)=9 \frac{x-y}{2}
\end{aligned}
$$

Since $A$ and $B$ are integers, $\frac{x+y}{2}$ and $\frac{x-y}{2}$ are also integers. If we denote these values by $a$ and $b$ respectively, we get:

$$
A=11 a, \quad B=9 b, \quad x=a+b, \quad y=a-b
$$

We need to determine the number of pairs $(a, b)$ for which $A$ and $B$ are two-digit numbers, and $x$ and $y$ are non-zero digits. Thus, $1<b<a$ and $a+b \leq 9$. If $a+b=9$, then the pairs $(7 ; 2)$, $(6 ; 3)$, $(5 ; 4)$ satisfy the condition. If $a+b=8$, then the pairs $(6 ; 2)$, $(5 ; 3)$ satisfy the condition. If $a+b=7$, then the pairs $(5 ; 2)$, $(4 ; 3)$ satisfy the condition. If $a+b=6$, then the pair $(4 ; 2)$ satisfies the condition. Finally, if $a+b=5$, then the pair $(3 ; 2)$ satisfies the condition. From these, we get the following 9 solutions:

| $A$ | 77 | 66 | 55 | 66 | 55 | 55 | 44 | 44 | 33 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $B$ | 18 | 27 | 36 | 18 | 27 | 18 | 27 | 18 | 18 |