Japan Mathematical Olympiad

$\frac{100}{495001}$

For any positive real numbers $x, y$, and $z$, the following equation holds:
$$(x \star y) \star z = \frac{x \star y}{(x \star y)z + 1} = \frac{\frac{x}{xy+1}}{\frac{x}{xy+1}z + 1} = \frac{x}{xy + xz + 1} = \frac{x}{x(y+z) + 1} = x \star (y+z).$$
Using this repeatedly, we have:
$$\begin{aligned} & ((\cdots((((100 \star 99) \star 98) \star 97) \star 96) \star \cdots \star 3) \star 2) \star 1 \\ &= ((\cdots((((100 \star (99 + 98)) \star 97) \star 96) \star \cdots \star 3) \star 2) \star 1 \\ &= ((\cdots((100 \star (99 + 98 + 97)) \star 96) \star \cdots \star 3) \star 2) \star 1 \\ &= \cdots \\ &= 100 \star (99 + 98 + 97 + \cdots + 3 + 2 + 1) \\ &= 100 \star \frac{99 \cdot 100}{2}. \end{aligned}$$
$$\text{Therefore, the answer is } 100 \star 4950 = \frac{100}{495001}.$$