10. A pair of positive integers is called "compatible" if and only if one of the numbers is the sum of the digits of the two numbers, and the other number is the product of the digits of the two numbers. Try to find:
(1) all compatible pairs of positive integers within 100;
(2) all compatible pairs of positive integers within 1000, where one of the numbers is greater than 99.
10. (1) Let $s, p$ be a compatible pair of numbers, where $s$ is their sum and $p$ is their product. Since $p \neq 0$, all the digits are non-zero.
We will discuss this in cases.
(i) If $s, p$ include a single-digit number, then $s > p$.
This indicates that $p$ is a single-digit number. It is easy to see that $s$ cannot be a single-digit number. Since $p$ is the product of the digits, we have $s = 11$, and thus, $p = 9$.
(ii) If both $s, p$ are two-digit numbers, we can set
$$
s = 10a + b, \quad p = 10c + d,
$$
where $a, b, c, d$ are all non-zero digits, then
$$
\left\{\begin{array}{l}
10a + b = a + b + c + d, \\
10c + d = abcd.
\end{array}\right.
$$
From equation (1), we have $9a = c + d$.
Thus, $a = 1$ or 2.
When $a = 2$, $c = d = 9$, substituting into equation (2) gives $99 = 162b$, which has no integer solution.
When $a = 1$, $c + d = 9$. From equation (2), we get
$9 + 9c = bc(9 - c)$
$\Rightarrow c \mid 9 \Rightarrow c = 1, 3$ or 9.
If $c = 1$, then $18 = 8b$, which is impossible.
Similarly, $c = 9$ is also impossible.
If $c = 3$, we get $b = 2$, at this point, $s = 12, p = 36$.
Therefore, within 100, there are only two pairs of compatible positive integers, which are $\{9, 11\}$ and $\{12, 36\}$.
(2) From the discussion in (1), we know that all the digits are non-zero, and $s$ cannot be a single-digit number.

Notice that the sum of six digits is at most 54, so $s$ can only be a two-digit number. Thus, $p$ is a three-digit number.

Let $s = 10a + b, p = 100c + 10d + e$, where $a, b, c, d, e$ are all non-zero digits. Then
$$
\left\{\begin{array}{l}
10a + b = a + b + c + d + e, \\
p = abcde.
\end{array}\right.
$$
Thus, $9a = c + d + e$
$\Rightarrow c + d + e = 9, 18$ or 27.
It is easy to verify that $c + d + e = 27$ is impossible.
Notice that the sum of the digits of $p$ is a multiple of 9.
Thus, $9 \mid p$.
(i) If $c + d + e = 18$, then
$a = 2, p = 2bcde$.
Thus, $p$ is even. Therefore, $e$ is also even.
Thus, $4 \mid p$.
Therefore, $p = 36n$.
From $100 \leqslant p \leqslant 999 \Rightarrow 3 \leqslant n \leqslant 27$.
Also, since $p$ is the product of the digits, the prime factors of $p$ can only be single-digit numbers, so
$$
n \neq 11, 13, 17, 19, 22, 23, 26.
$$
After verification, among the remaining numbers, only when
$$
n = 8, 16, 18, 21, 24, 27
$$
does $p = 36n$ have a digit sum of 18. It is easy to verify that none of these numbers satisfy the condition.
(ii) If $c + d + e = 9$, then
$a = 1, p = bcde$.
By the AM-GM inequality, we know $cde \leqslant 27$.
Thus, $p = 9n, 12 \leqslant n \leqslant 27$.
Similar to (i), we have
$n \neq 13, 17, 19, 22, 23, 26$.
Checking each one, $n = 15, 16$ satisfy the condition.
When $n = 15$, $p = 135, b = 9$;
When $n = 16$, $p = 144, b = 9$.
In summary, the required pairs of compatible positive integers are $\{135, 19\}$ and $\{144, 19\}$.