Why do I love 2015 so much? 3

Its easy to see that one solution to the equation is $x=100 , y=1$ .Since its a Diophantine equation , we get general solutions for $x,y$ as follows:

$\forall k \in \mathbb{Z} \begin{cases} x=100-3k \\ y=4k+1 \end{cases}$

But since we want positive integer solutions , we have the bound $0\leq k \leq 33$ . Thus we have solutions in arithmetic progression for x,y:

$x=100,97,94 \dots 4,1 \\ y=1,5,9 \dots 129,133$

And by using formula for sum of Arithmetic progression yields the sum as $\boxed{3995}$ .

$\begin{aligned} 20x+15y & = 2015 \\ \Rightarrow 4x + 3y & = 403 \\ \color{#3D99F6}{3(x+y)} + x & = \color{#3D99F6}{402} + 1 \quad \quad \small \color{#3D99F6}{\text{Note that 402 is divisible by 3.}} \end{aligned}$

The above shows that $(x, y) = (1,133)$ is a solution pair. Since $3(x+y)$ must be a multiple of $3$ therefore, $x$ is of the form $x = 3\color{#3D99F6}{k} + 1$ , where $\color{#3D99F6}{k} = 0,1,2,3,...$ , and:

$\Rightarrow y = \dfrac{402-3k}{3} - x = 134 - k - 3k - 1 = 133-4k$

Since $y > 0 \quad \Rightarrow k < \dfrac{133}{4} \quad \Rightarrow k = 0,... 33$

Therefore, the sum of all $x$ and $y$ :

$\begin{aligned} \sum_{k=0}^{33} (x+y) = \sum_{k=0}^{33} (3k+1 + 133 - 4k) = \sum_{k=0}^{33} (134 - k) = \boxed{3995} \end{aligned}$