10.2. Determine the greatest possible value of the ratio between the sum of the digits of a four-digit number and the number itself.
Solution: Let $\overline{a b c d}$ be a four-digit number. Then

$$
\begin{gathered}
\frac{\overline{a b c d}}{a+b+c+d}=\frac{1000 a+100 b+10 c+d}{a+b+c+d}=1+\frac{999 a+99 b+9 c}{a+b+c+d} \\
\text { Since } d \leq 9, \text { we get } \frac{\overline{a b c d}}{a+b+c+d} \geq 1+\frac{999 a+99 b+9 c}{a+b+c+9}= \\
=1+\frac{999(a+b+c+9)-900 b-990 c-999 \cdot 9}{a+b+c+9}=1+999-\frac{900 b+990 c+999 \cdot 9}{a+b+c+9}
\end{gathered}
$$

Since $a \geq 1$, we get $\frac{\overline{a b c d}}{a+b+c+d} \geq 1000-\frac{900 b+990 c+999 \cdot 9}{b+c+10}=$ $=1000-\frac{900(b+c+10)+90 c-9}{b+c+10}=1000-900-\frac{90 c-9}{b+c+10}=100-\frac{90 c-9}{b+c+10}$.

Since $b \geq 0$, we get $\frac{\overline{a b c d}}{a+b+c+d} \geq 100-\frac{90 c-9}{c+10}=100-\frac{90(c+10)-909}{c+10}=$ $=100-90+\frac{909}{c+10}=10+\frac{909}{c+10}$.

Since $c \leq 9$, we get $\frac{\overline{a b c d}}{a+b+c+d} \geq 10+\frac{909}{19}=\frac{1099}{19}$.

Therefore, $\frac{a+b+c+d}{\overline{a b c d}} \leq \frac{19}{1099}$, so the maximum value is $\frac{19}{1099}$ for the number 1099. Answer: $\frac{19}{1099}$.