$10 \cdot 44$ What is the maximum difference between a three-digit number and the sum of the cubes of its digits? What kind of three-digit number can achieve this maximum difference? What is the smallest positive difference?

(10th All-Russian Mathematical Olympiad, 1984)
[Solution] Let $abc$ be any three-digit number, where $a \in \{1,2, \cdots, 9\}, b, c \in \{0,1,2, \cdots, 9\}$.
Let $f(\overline{abc}) = \overline{abc} - (a^3 + b^3 + c^3)$, then
$$
\begin{array}{l}
f(\overline{abc}) = 100a + 10b + c - (a^3 + b^3 + c^3) \\
= (100a - a^3) + (10b - b^3) + (c - c^3) . \\
= f_1(a) + f_2(b) + f_3(c) .
\end{array}
$$

where $f_1(a) = 100a - a^3, \quad f_2(b) = 10b - b^3, \quad f_3(c) = c - c^3$. The maximum value of $f(\overline{abc})$ is achieved when and only when $f_1(a), f_2(b), f_3(c)$ all take their maximum values. Now we calculate $f_1(a)$.
$$
\begin{array}{lll}
f_1(1) = 99, & f_1(2) = 192, & f_1(3) = 273, \\
f_1(4) = 336, & f_1(5) = 375, & f_1(6) = 384, \\
f_1(7) = 357, & f_1(8) = 288, & f_1(9) = 171 .
\end{array}
$$

From the above, we see that when $a = 6$, $f_1(a) = 384$ is the maximum.
$$
\begin{array}{l}
f_2(0) = 0, \quad f_2(1) = 9, \quad f_2(2) = 12, \quad f_2(3) = 3, \\
f_2(4) = -24 < 0, \quad f_2(b) < 0 \quad (b = 4,5, \cdots, 9)
\end{array}
$$

From the above, we see that when $b = 2$, $f_2(b) = 12$ is the maximum.
Clearly, when $c \geq 2$, $f_3(c) < 0$. And
$$
f_3(0) = 0, \quad f_3(1) = 0 .
$$

Therefore, when $\overline{abc} = 620$ or 621, $f(\overline{abc})$ achieves its maximum value, which is
$$
384 + 12 + 0 = 396 .
$$

Next, consider the minimum positive value of $f(\overline{abc})$. Since
$$
\begin{aligned}
f_1(a) & = 100a - a^3 \\
& = 99a + (a - a^3) \\
& = 99a - (a-1)a(a+1), \\
f_2(b) & = 10b - b^3 \\
& = 9b + (b - b^3) \\
& = 9b - (b-1)b(b+1), \\
f_3(c) & = c - c^3 \\
& = -(c-1)c(c+1) .
\end{aligned}
$$

Since the product of three consecutive integers is divisible by 3, $f_1(a), f_2(b)$, and $f_3(c)$ are all divisible by 3, and thus $f(\overline{abc})$ is also divisible by 3.
Therefore, the minimum positive value of $f(\overline{abc}) \geq 3$.
It can be verified that:
$$
f(437) = f(474) = f(856) = 3 .
$$

So the minimum positive value of $f(\overline{abc})$ is 3.