, 2022

Solution:
First consider the case when $abc = def + 1$. Let $X = abc$. Then
$$\frac{abc - def}{ghi} = \frac{1}{ghi} = \frac{abcdef}{9!} = \frac{X \cdot (X-1)}{9!}$$
Because $X > 1$, this is an increasing function of $X$. Note that $X(X - 1) = abcdef \ge 6!$, and therefore $X \ge 28$. Either $X$ or $X - 1$ is odd and is therefore a product of three distinct odd factors from $\{1, 2, \dots, 9\}$.
If $X - 1 = 27$, then $\{d, e, f\} = \{1, 3, 9\}$. However, $X = 28$ is not factorable into three distinct positive integers each greater than $1$. The next odd number greater than $27$ whose prime factors are less than $10$ is $35 = 7 \cdot 5 \cdot 1$. If either $X$ or $X - 1$ is $35$, then the other is either $34$ or $36$. But $17$ is a prime factor of $34$, so this forces $X$ to be $36 = 6 \cdot 3 \cdot 2$.
Therefore in the case when $abc$ and $def$ are consecutive integers, the minimum value of the expression is
$$\frac{36 \cdot 35}{9!} = \frac{6 \cdot 3 \cdot 2 - 7 \cdot 5 \cdot 1}{9 \cdot 8 \cdot 4} = \frac{1}{288}$$
If $abc \ge def + 2$, then
$$\frac{abc - def}{ghi} \ge \frac{2}{9 \cdot 8 \cdot 7} = \frac{1}{252} > \frac{1}{288}$$
Hence the least possible positive value of the expression is $\frac{1}{288}$. The requested sum is $1 + 288 = 289$.