Solution:

Shift the graph down by $2$. The problem is equivalent to finding the smallest slope $m\gt\dfrac12$ for which $y=mx$ passes through a lattice point with $0\lt x\le100$.

For a fixed integer $x$, the smallest integer $y$ with $y/x\gt1/2$ is $x/2+1$ when $x$ is even, and $(x+1)/2$ when $x$ is odd.

Thus the candidate slopes are $\dfrac12+\dfrac1x$ for even $x$, minimized at $x=100$ as $\dfrac{51}{100}$, and $\dfrac12+\dfrac1{2x}$ for odd $x$, minimized at $x=99$ as $\dfrac{50}{99}$.

The smaller of these is $\dfrac{50}{99}$, so every $m$ with $\dfrac12\lt m\lt\dfrac{50}{99}$ avoids such lattice points, and this upper endpoint is best possible.

Thus, B is the correct answer.