Answer: 85

Small Hint:

Write $x=ay$ and reduce $x+1$ modulo $y+1$

Big Hint:

The quotient $a$ must have the form $1+t(y+1)$; count the possible positive values of $t$

Solution:

Write $x=ay.$ Modulo $y+1,$ we have $y\equiv-1,$ so $x+1$ is divisible by $y+1$ exactly when $a\equiv1\pmod {y+1}.$ Since $x>y,$ write $a=1+t(y+1)$ with $t\geq1.$ The bound $x\leq100$ becomes $$t\leq\left\lfloor\frac{100-y}{y(y+1)}\right\rfloor.$$ Only $1\leq y\leq9$ contribute, giving $$\begin{aligned}49+16+8&+4+3\\&+2+1+1+1\\&=85.\end{aligned}$$