Japan Mathematical Olympiad Initial Round

$$\boxed{\frac{3}{16\sqrt{7}}}$$
From $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$ it follows that there are $(8+1) \cdot (4+1) \cdot (2+1) \cdot (1+1) = 270$ positive factors of $10!$. Suppose we enumerate them as $d_1, d_2, \dots, d_{270}$ in increasing order of magnitude, starting with the smallest one as $d_1$, then for each $k$, $1 \le k \le 270$, we have
$$d_k \cdot d_{271-k} = 10!,$$
and therefore, we get
$$\frac{1}{d_k + \sqrt{10!}} + \frac{1}{d_{271-k} + \sqrt{10!}} = \frac{d_k + d_{271-k} + 2\sqrt{10!}}{\sqrt{10!}(d_k + d_{271-k}) + 2 \cdot 10!} = \frac{1}{\sqrt{10!}}$$
Consequently, the desired sum is given by
$$\frac{1}{2} \sum_{k=1}^{270} \left( \frac{1}{d_k + \sqrt{10!}} + \frac{1}{d_{271-k} + \sqrt{10!}} \right) = \frac{1}{2} \cdot 270 \cdot \frac{1}{\sqrt{10!}} = \frac{3}{16\sqrt{7}}$$