Aim for a numerator of $1:$ look for two triples of distinct digits whose products differ by exactly $1$

Save the three largest leftover digits for the denominator, then rule out bigger denominators by showing they force a numerator difference of at least $2$

Try to make the numerator equal to $1$ while keeping large digits in the denominator. The products $2 \cdot 3 \cdot 6 = 36$ and $1 \cdot 5 \cdot 7 = 35$ differ by $1$ and leave $4, 8, 9$ for the denominator, giving the value $$\frac{36 - 35}{4 \cdot 8 \cdot 9} = \frac{1}{288}.$$

To beat this, a fraction would need numerator $1$ with denominator greater than $288.$ The denominators exceeding $288$ are $504,$ $432,$ $360,$ $378,$ $315,$ and $336,$ coming respectively from $\{7,8,9\},$ $\{6,8,9\},$ $\{5,8,9\},$ $\{6,7,9\},$ $\{5,7,9\},$ and $\{6,7,8\}.$ Splitting the remaining six digits into two triples, the smallest positive numerator differences are respectively $6,$ $2,$ $8,$ $2,$ $4,$ and $6.$ The resulting lower bounds $$\begin{aligned}&\frac{6}{504},\quad \frac{2}{432},\quad \frac{8}{360},\\&\frac{2}{378},\quad \frac{4}{315},\quad \frac{6}{336}\end{aligned}$$ all exceed $\frac{1}{288}.$

So the minimum positive value is $\frac{1}{288},$ and $m + n = 1 + 288 = 289.$