Let's divide every natural number with at least two digits by the sum of its digits! When will the quotient be the largest, and when will it be the smallest?
Among the quotients, there is no largest, because among them are the quotients obtained from numbers of the form $10^{k}$, which are equal to $10^{k}$ $(k>1)$. When looking for the smallest quotient, let's first examine the two-digit numbers:

$$
h=\frac{10 a+b}{a+b}=1+\frac{9}{1+b / a}
$$

thus, the quotient is the smallest when $b / a$ is the largest, which occurs when $b=9, a=1$, in which case $h=1.9$. Let $A$ be any $n$-digit number, where $n \geq 3$. Then $A \geq 10^{n-1}$, and the sum of its digits is at most $9 n$, so the quotient is at least $10^{n-1} / 9 n$. This is already greater than 1.9 when $n=3$ and increases monotonically with $n$:

$$
\frac{10^{n}}{9(n+1)}>\frac{10^{n-1}}{9 n}
$$

since this inequality is equivalent to $10>1+\frac{1}{n}$. Therefore, the smallest quotient is 1.9.