Example 3 Find all positive integers $m$ such that there exist pairwise coprime positive integers $a, b, c$ satisfying
$$
S(ab)=S(ac)=S(bc)=m .
$$
【Analysis】This problem seems difficult to approach, but it is related to the sum of digits, and generally, we would try modulo 3 first. It can be observed that $m$ modulo 3 should leave a remainder of 1, and the construction method is the same as Example 2.

Solution Assume $a b c$ is a multiple of 3. Then $m$ is a multiple of 3.

Also, $a, b, c$ are pairwise coprime, so two of $a, b, c$ are not multiples of 3, and the product of these two numbers is not a multiple of 3. However, $m$ is the sum of the digits of the product, which leads to a contradiction. Therefore, $a b c$ is not a multiple of 3, and thus, $a, b, c$ are not multiples of 3. They must have two numbers with the same remainder when divided by 3, and the product of these two numbers leaves a remainder of 1 when divided by 3. Therefore, all $m$ leave a remainder of 1 when divided by 3.
(1) When $a=b=c=1$, we have
$S(a b)=S(a c)=S(b c)=1$;
When $a=8, b=\underset{k \uparrow}{89 \cdots 9}, c=\underset{k-1 \uparrow}{19 \cdots 97}(k \geqslant 1)$, we have
$$
S(a b)=S(a c)=S(b c)=9 k+1,
$$

and $a, b, c$ are pairwise coprime.
(2) When $a=2, b=11, c=101$, we have
$S(a b)=S(a c)=S(b c)=4$;
When $a=2, b=\underset{k \uparrow}{22 \cdots 9}, c=\underset{k-1 \uparrow}{49 \cdots 97}(k \geqslant 1)$, we have
$$
S(a b)=S(a c)=S(b c)=9 k+4,
$$

and $a, b, c$ are pairwise coprime.
(3) When $a=4, b=13, c=31$, we have
$S(a b)=S(a c)=S(b c)=7$;
When $a=4, b=\underset{k \uparrow}{49 \cdots 9}, c=\underset{k-1 \uparrow}{69 \cdots 9}(k \geqslant 1)$, we have
$S(a b)=S(a c)=S(b c)=9 k+7$,

and $a, b, c$ are pairwise coprime.
In summary, $m$ is all positive integers that leave a remainder of 1 when divided by 3.