The smallest positive integer $n$for which $\sqrt{100+\sqrt{\mathrm{n}}}+\sqrt{100-\sqrt{\mathrm{n}}}$  is an integer, is $k$then sum of digits of $k$ is

The number $\sqrt{100+\sqrt{\mathrm{n}}}+\sqrt{100-\sqrt{\mathrm{n}}}$  is  a positive integer if and only if its square is a perfect square. We have
${\left(\sqrt{100+\sqrt{\mathrm{n}}}+\sqrt{100-\sqrt{\mathrm{n}}}\right)}^2=\left(100+\sqrt{\mathrm{n}}\right)+\left(100-\sqrt{\mathrm{n}}\right)+2\sqrt{\left(100+\sqrt{\mathrm{n}}\right)+\left(100-\sqrt{\mathrm{n}}\right)}$
$=200+2\sqrt{10000-\mathrm{n}}.$
To minimize n, we should maximize the value of this expression, given that it is a perfect square. For this expression to be a perfect square, $\sqrt{10000-\mathrm{n}}$  must be an integer. Then   is even, and it $200+2\sqrt{10000-\mathrm{n}}$ is less than  $200+2\sqrt{10000}=400={20}^2.$ Therefore, the greatest possible perfect square value of $200+2\sqrt{10000-\mathrm{n}}$  is ${18}^2=324.$  Solving 
$200+2\sqrt{10000-\mathrm{n}}=324$
For n gives the answer,   $\mathrm{n}=6156.$