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 $\{7,8,9\},$ $\{6,8,9\},$ $\{5,8,9\},$ $\{6,7,9\},$ $\{5,7,9\},$ and $\{6,7,8\}.$ Splitting the remaining six digits into two triples in each case, the closest product pairs are $30$ and $24,$ $30$ and $28,$ $36$ and $28,$ $30$ and $28,$ $36$ and $32,$ and $36$ and $30,$ respectively — differences of at least $2,$ and even $\frac{2}{432} = \frac{1}{216}$ exceeds $\frac{1}{288}.$

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