First of all note that $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$ hence

- $10!$ is not a perfect square,
- $10!$ has $(8+1)(4+1)(2+1)(1+1) = 270$ divisors.

If $d \mid 10!$ , then there exists $p$ such that $dp =10!$ meaning that $p$ is also a divisor of $10!$. We also note that if one of $p, q$ is less than $10!$ then the other is greater than $10!$. Note that the case $p=q$ is impossible since $10!$ is not a perfect square.

Taking into account the addent that corresponds to $d$ along with the addent that corresponds to $p$ , we get:

$$\frac{1}{d+\sqrt{10!}} + \frac{1}{p+\sqrt{10!}}= \frac{1}{d+\sqrt{10!}} + \frac{1}{ \frac {10!}{d}+\sqrt{10!}} = \frac{1}{d+\sqrt{10!}} + \frac{d}{\sqrt{10!}( \sqrt{10!}+d)} = \frac{1}{\sqrt{10!}}$$

which is constant. Since we have $\frac{270}{2} = 135$ such pairs , we conclude that

$$\sum_{{\rm d}\mid 10!}\frac{1}{{\rm d}+\sqrt{10!}} = \frac{135}{\sqrt{10!}} = \frac{3 \sqrt{7}}{112}$$