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A quick overview of a solution ....

Clearly $m=1$ works, so now look at two-digit integers $ab$ .

We require that $b∗(a+b)=10a+b⟹b∗(a+b-1)=10a$ . Testing out combinations of $a,b$ gives us $45$ and $24$ as the only possible numbers, and so $45$ is the largest such two-digit number. With $3$ -digit numbers $abc$ the largest possible product is $9∗27=243$ , so we only need to look at the cases where $a=1$ and $a=2$ . Thus we need to look for any solutions to

$c∗(b+c)=10∗(b+10)$ and $c∗(b+c+1)=10∗(b+20)$ ,

where each of $b$ and $c$ are between $0$ and $9$ inclusive.

There are no such solutions to either equation.

Since $9∗36=324$ there can be no such $4$ -digit integers, and in fact there can be no $k$ -digit integers, $k≥4$ , that meet the requirements. Thus $m = 45$ .

Looking now for $n$ , a similar approach reveals that $48$ is the largest $2$ -digit integer. Now again since $9*27 = 243$ , the largest potential $3$ -digit integer will have a leading digit of $2$ , yielding a largest possible product of the leading digit and the sum of the digits of $2*(2 + 3 + 9) = 28$ , i.e., a two-digit number.

Thus $n = 48$ , and so $n - m = 48 - 45 = \boxed{3}$ .