26. Answer: 48
Let $x=\overline{a b c d}$ and $y=\overline{d o b a}$ where $a, d \neq 0$. Then
$$
\begin{aligned}
y-x & =1000 \times d-d+100 \times c-10 \times c+10 \times b-100 \times b+a-1000 \times a \\
& =999(d-a)+90(c-b)=9(111(d-a)+10(c-b)) .
\end{aligned}
$$

So we have $111(d-a)+10(c-b)=353$. Consider the remainder modulo 10 , we obtain $d-a=3$, which implies that $c-b=2$. Thus the values of $a$ and $b$ determines the values of $d$ and $c$ respectively.
$a$ can take on any value from 1 to 6 , and $b$ can take any value from 0 to 7 , giving $6 \times 8=48$ choices.
26. Given any 4-digit positive integer $x$ not ending in ' 0 ', we can reverse the digits to obtain another 4 -digit integer $y$. For example if $x$ is 1234 then $y$ is 4321 . How many possible 4-digit integers $x$ are there if $y-x=3177$ ?