1. Total number of pairs of positive integers

Let the two positive integers be $x$ and $y$ such that $x+y=24,$ with $x,y\in \mathbb{Z}^+$.

The number of positive integer solutions is: $x=1,2,3,\dots,23$ with corresponding $y=23,22,21,\dots,1$.

So total ordered pairs $=23$.

2. Greatest possible product

For a fixed sum, the product is maximum when the numbers are as close as possible.

Since $x+y=24,$ the maximum product occurs at $x=y=12.$ Thus, $P_{\max}=12\cdot 12=144.$

Therefore, $\frac{3}{4}P_{\max}=\frac{3}{4}\cdot 144=108.$

We need $xy\ge 108.$

3. Find all positive integer pairs satisfying $x+y=24$ and $xy\ge 108$

Put $y=24-x$. Then $x(24-x)\ge 108.$

So, $24x-x^2\ge 108$ $x^2-24x+108\le 0.$

Factorizing, $(x-6)(x-18)\le 0.$

Hence, $6\le x\le 18.$

Since $x$ is an integer, favorable values are $x=6,7,8,\dots,18,$ which gives $18-6+1=13$ favorable ordered pairs.

4. Probability

Thus, $\text{Probability}=\frac{13}{23}.$

So, $m=13,\quad n=23.$

Hence, $n-m=23-13=10.$

5. Compare with stored answer

Our derived answer is 10, which matches option A.