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 $$\frac{1}{2}\sum _{k=1}^{270}\left(\frac{1}{d_k+\sqrt{10!}}+\frac{1}{d_{271-k}+\sqrt{10!}}\right)=\frac{1}{2}\cdot 270\cdot \frac{1}{\sqrt{10!}}=\frac{3}{16\sqrt{7}}$$