Pairs...

Lets assume that x and y are natural numbers. Then,
$2015\quad \equiv \quad 3x\quad (mod\quad 2)$ .Note that every odd 3x would do. The possible values of x are 1,3,5..................................671. Now calculate number of terms to get answer $\boxed{336}$ .

Now I assume that $\ x \ and \ y$ are natural numbers. Using odd+even = odd, means that there are solutions (x,y), if and only if , x is odd and satisfies $3x \leq\ 2013 \Rightarrow\ x \leq\ 671 \ \ , \ \forall x \in N$ Now all numbers are of the form 2n-1, so we create a simple linear equation for ${n}$ : $2n-1 = 671 \therefore\ n = 336$

Doing some modulo arithmetic we get:

$3x+2y \equiv 2015 \quad (mod \quad 2)$

$x \equiv 1 \quad (mod \quad 2) \Rightarrow x=2k+1$

Replacing the value found for $x$ we get:

$6k+3+2y=2015 \rightarrow 2y=2012-6k \rightarrow y=1006 - 3k \rightarrow$

$\quad since \quad y \quad cannot \quad equal \quad 0 \quad : \quad 3k \lt 1006$

$k \lt \lfloor \frac {1006}{3} \rfloor \rightarrow k\le 335 \rightarrow$

$\quad \quad \quad 0 \le k \le 335 \rightarrow k \quad can \quad have \quad 335+1=336 \quad values.$