Solution:

For a fixed slope $m$, the number of points in $S$ on or below $y=mx$ is

$$\sum_{x=1}^{30}\lfloor mx\rfloor,$$

for the slopes near the answer.

At $m=\frac23$, grouping $x=3k+1,3k+2,3k+3$ for $k=0,1,\ldots,9$ gives

$$\lfloor 2x/3\rfloor=2k,\ 2k+1, \ 2k+2,$$

whose sum over each block is $6k+3$. Thus the total is

$$\sum_{k=0}^9(6k+3)=270+30=300.$$

If $m<\frac23$, the ten points with ratios $y/x=2/3$ are no longer counted, so the count is less than $300$. Therefore the lower end is $\frac23$.

The next possible ratio $y/x$ greater than $\frac23$, with $1\le x,y\le30$, is minimized by checking $x$ modulo $3$. The best candidates are

$$\frac{19}{28},\qquad \frac{20}{29},\qquad \frac{21}{30}=\frac{7}{10},$$

and the smallest is $\frac{19}{28}$. Hence the interval length is

$$\frac{19}{28}-\frac23=\frac1{84}.$$

Thus $a+b=1+84=85$.

Thus, the answer is E .