jmc

Since $y|x$, $y+1|x+1$, then $\text{gcd}(y,x)=y$ (the bars indicate divisibility) and $\text{gcd}(y+1,x+1)=y+1$. By the Euclidean algorithm, these can be rewritten respectively as $\text{gcd}(y,x-y)=y$ and $\text{gcd}(y+1,x-y)=y+1$, which implies that both $y,y+1|x-y$. Also, as $\text{gcd}(y,y+1)=1$, it follows that $y(y+1)|x-y$. [1] Thus, for a given value of $y$, we need the number of multiples of $y(y+1)$ from $0$ to $100-y$ (as $x\le 100$). It follows that there are $\lfloor \frac{100-y}{y(y+1)}\rfloor$ satisfactory positive integers for all integers $y\le 100$. The answer is $$\sum _{y=1}^{99}\lfloor \frac{100-y}{y(y+1)}\rfloor =49+16+8+4+3+2+1+1+1=\boxed{85}.$$