What is the greatest result we can find when we divide a two-digit number by the sum of its digits?
We are looking for the maximum value of $\frac{10 a+b}{a+b}$, where $a$ and $b$ represent digits, with at least one different from 0. We have

$$
\frac{10 a+b}{a+b}=\frac{10 a+10 b-9 b}{a+b}=\frac{10 a+10 b}{a+b}-\frac{9 b}{a+b}=10-\frac{9 b}{a+b} \leq 10
$$

Therefore, if we can find $a$ and $b$ such that $\frac{10 a+b}{a+b}=10$, we will have the maximum result. Note that this occurs when $b=0$, that is:

$$
\frac{10}{1}=\frac{20}{2}=\frac{30}{3}=\frac{40}{4}=\frac{50}{5}=\frac{60}{6}=\frac{70}{7}=\frac{80}{8}=\frac{90}{9}=10 .
$$

Therefore, the answer is 10.