Looks hard

Assum $d=gcd(a,b)$ . Then

$(a-b)\sqrt{ab}=d^2\cdot (a'-b')\sqrt{a'b'}=2^5\cdot 3^2\cdot 7$

Note that $a=a'\cdot d \ ,\ b=b'\cdot d$ , such that $gcd(a',b')=1$ . We can take $a'=A^2 \ , \ b'=B^2$ , since $\sqrt{a'b'}$ should be a whole number. consequently, All the terms in the product $(a'-b')\sqrt{a'b'}= A^2-B^2 \sqrt{A^2 \cdot B^2}=(A+B)\cdot A \cdot B \cdot (A-B)$ are relatively prime, except for $gcd(A-B,A+B)$ that can either $1$ or $2$ . If we re-write the original equation

$d^2(A+B)\cdot A \cdot B \cdot (A-B)=2^5\cdot 3^2\cdot 7$

Now, we need to partition $\frac{1}{d^2}\cdot 2^5\cdot 3^2\cdot 7$ into $4$ groups (integer $1$ is a legitimate partition), such that they meet the mentioned requirements.

if $d=1$ then

$(A+B,A,B,A-B)=(2^4,3^2,7,2) \implies a=d \cdot A^2 =81$

is the only solution

if $d=2$ then

$(A+B)\cdot A \cdot B \cdot (A-B)=2^3\cdot 3^2\cdot 7$

and

$(A+B,A,B,A-B)=(3^2,2^3,1,7) \implies a=d \cdot A^2 =128$

is the only solution.

For other possible $d$ 's, there would be no solutions.

The sum of the valid solutions is $81+128=209$

Given that $(a-b)\sqrt{ab} = 2016$ , for positive integers $a$ and $b$ , solutions exist only when $a > b$ . Let $a=b+c$ , where $c$ is also a positive integer. Then we have

$\begin{aligned} c\sqrt{b(b+c)} & = 2016 \\ \sqrt{b(b+c)} & = \frac {2016}c & \small \blue{\text{For both sides to be integers, }c \text{ must be a divisor of 2016.}} \\ b^2 + cb - \frac {2016^2}{c^2} & = 0 \\ \implies b & = \frac {-c + \sqrt{c^2 + 4\left(\frac{2016^2}{c^2}\right)}}2 \end{aligned}$

For $b$ to be an integer, the sum $c^2 + \dfrac {4(2016^2)}{c^2}$ must be a perfect square. Since $2016 = 2^5\cdot 3^2 \cdot 7$ and $c$ is its divisor, there are $(5+1)(2+1)(1+1) = 36$ values of $c$ . Out of the $36$ values of $c$ , the sum is perfect square when,

$\begin{cases} \begin{array}{lll} c = 1 & \implies \red{b = 2015.5} & \small \red{b \text{ is not an integer, rejected.}} \\ c = 32 & \implies b = 49 & \implies a = 81 \\ c = 126 & \implies b = 2 & \implies a = 128 \end{array} \end{cases}$

Therefore the sum of all values of $a$ , $81+128=\boxed{209}$ .