Difficult sets of Equations, counting 
Let $ m$ be the number of solutions in positive integers to the equation $ 4x+3y+2z=2009$, and let $ n$ be the number of solutions in positive integers to the equation $ 4x+3y+2z=2000$. Find the remainder when $ m-n$ is divided by $ 1000$.

Please no complete solutions.
$4x + 3y + 2z = 2009$ first.
$2z = 2009 - 4x - 3y$
$\implies 2z \equiv 2 - 4x \pmod{3} \implies z \equiv 1 - 2x \pmod{3}$.So $z = 1 - 2x + 3k$. 
$4x + 3y + 2 - 4x +  6k = 2009$.
$3y = 2007 + 6k$.
$ y = (2007 + 6k)/3 = 2007/3 + 2k = 669 + 2k$.
But that doesnt work properly. 
I literally cant do ANYTHING, I am very stuck!
EDIT:
$x, y, z = 1$ each.
$3y + 2z = 2005 \implies 3y \equiv 1 \equiv 3 \pmod{2} \implies y = 1 + 2k, z = \frac{2005 - 3(1 + 2k)}{2} = 1001 - 3k.$
$2x + z = 1003 \implies z \equiv 1 \pmod{2}, z = 1 + 2k, x = \frac{1002 - 2k}{2} = 501 - k$
$4x + 3y = 2007 \implies 3y \equiv 3 \pmod{4} \implies y = 1 + 4k, x = \frac{2007 - 3y}{4} = 1002 - 3k$
I just cannot find how many $k$ values will work. 

If $(x,y,z)=(a,b,c)$ is a solution of $4x+3y+2z=2000$, then $(x,y,z)=(a+1,b+1,c+1)$ is a solution of $4x+3y+2z=2009.$
Now find the number of the solutions $(x,y,z)$ of $4x+3y+2z=2009$ such that either $x,y$ or $z$ equals $1$. (note that this is $m-n$.)
