smc

First, we find a numerical representation for the number of lattice points in $S$ that are under the line $y=mx.$ For any value of $x,$ the highest lattice point under $y=mx$ is $\lfloor mx\rfloor .$ Because every lattice point from $(x,1)$ to $(x,\lfloor mx\rfloor )$ is under the line, the total number of lattice points under the line is ${\sum }_{x=1}^{30}(\lfloor mx\rfloor ).$ Now, we proceed by finding lower and upper bounds for $m.$ To find the lower bound, we start with an approximation. If $300$ lattice points are below the line, then around $\frac{1}{3}$ of the area formed by $S$ is under the line. By using the formula for a triangle's area, we find that when $x=30,y\approx 20.$ Solving for $m$ assuming that $(30,20)$ is a point on the line, we get $m=\frac{2}{3}.$ Plugging in $m$ to ${\sum }_{x=1}^{30}(\lfloor mx\rfloor ),$ we get: $$\sum _{x=1}^{30}(\lfloor \frac{2}{3}x\rfloor )=0+1+2+2+3+\cdots +18+18+19+20$$ We have a repeat every $3$ values (every time $y=\frac{2}{3}x$ goes through a lattice point). Thus, we can use arithmetic sequences to calculate the value above: $$\sum _{x=1}^{30}(\lfloor \frac{2}{3}x\rfloor )=0+1+2+2+3+\cdots +18+18+19+20$$$$=\frac{20(21)}{2}+2+4+6+\cdots +18$$$$=210+\frac{20}{2}\cdot 9$$$$=300$$ This means that $\frac{2}{3}$ is a possible value of $m.$ Furthermore, it is the lower bound for $m.$ This is because $y=\frac{2}{3}x$ goes through many points (such as $(21,14)$). If $m$ was lower, $y=mx$ would no longer go through some of these points, and there would be less than $300$ lattice points under it. Now, we find an upper bound for $m.$ Imagine increasing $m$ slowly and rotating the line $y=mx,$ starting from the lower bound of $m=\frac{2}{3}.$The upper bound for $m$ occurs when $y=mx$ intersects a lattice point again (look at this problem to get a better idea of what's happening: https://artofproblemsolving.com/wiki/index.php/2011_AMC_10B_Problems/Problem_24). In other words, we are looking for the first $m>\frac{2}{3}$ that is expressible as a ratio of positive integers $\frac{p}{q}$ with $q\le 30.$ For each $q=1,\dots ,30$, the smallest multiple of $\frac{1}{q}$ which exceeds $\frac{2}{3}$ is $1,\frac{2}{2},\frac{3}{3},\frac{3}{4},\frac{4}{5},\cdots ,\frac{19}{27},\frac{19}{28},\frac{20}{29},\frac{21}{30}$ respectively, and the smallest of these is $\frac{19}{28}.$ Alternatively, see the method of finding upper bounds in solution 2. The lower bound is $\frac{2}{3}$ and the upper bound is $\frac{19}{28}.$ Their difference is $\frac{1}{84},$ so the answer is $1+84=\boxed{85}.$ An alternative would be using Farey fractions and the mediant theorem to find the upper bound. $\frac{2}{3}$ and $\frac{7}{10}$ gives $\frac{9}{13},$ and so on using Farey addition.