On existence of positive integer solution of $\binom{x+y}{2}=ax+by$ How can I prove this? 

Prove that for any two positive integers $a,b$ there are two positive integers $x,y$ satisfying the following equation:
  $$\binom{x+y}{2}=ax+by$$

My idea was that $\binom{x+y}{2}=\dfrac{x+2y-1}{2}+\dfrac{y(y-1)}{2}$ and choose $x,y$, such that $2a=x+2y-1, 2b=y(y-1)$, but using this idea, $x,y$ won’t be always positive. 

Case $a=b$ is trivial after dividing both sides by $x+y$, $a$ and $b$ are symmetric in the problem, so in the following text WLOG I assume $a < b$.
One can visualize the problem as follows. Denote $z=x+y$, then the left-hand side is $\frac{z(z-1)}{2}$. Let's subdivide the interval $(az, bz)$ into $z$ equal parts: then the right-hand side can be any point of subdivision (not including the ends) if you choose appropriate $x$ and $y$. So the problem is equivalent to finding $z$ such that $\frac{z(z-1)}{2}$ is a point of the subdivision of interval $(az, bz)$ into $z$ equal parts. If we multiply everything ($\frac{z(z-1)}{2}$, interval ends and points of the subdivision) in this condition by $2/z$, it becomes:
Condition. $(z-1)$ is a point of the subdivision of interval $(2a, 2b)$ into $z$ equal parts.
Now the simplest try of finding such $z$ seems to be: make all integer points between $2a$ and $2b$ the points of subdivision into $z$ equal parts, while $2a < z-1 < 2b$, making $z-1$ a point of such subdivision, because it is an integer. All integer points on $(2a, 2b)$ are in subdivision into $z$ equal parts iff $z$ is a multiple of $d=2b-2a$. Let's consider as possible values $(z-1)=-1,(-1+d),(-1+2d),(-1+3d),\cdots$ This is a progression with step $d$ which will definitely hit the segment $[2a, 2b]$ of length $d$, and because all its members are odd, it will actually hit the interval $(2a, 2b)$. The $z$ for which this happens satisfies both $2a < z-1 < 2b$ and $d \mid z$ conditions, as required.
