Number of solutions of: $3x+y=5702$ 
Find the number of ordered pairs $(x,y)$ satisfying  $3x+y=5702$ in
  natural numbers restricted by: $x+y\le2003$

I don't know any method for counting number of solutions of such equations...

I'll find all the integer solutions of $3x+y=5702$.
First, mod $3$ gives $y\equiv 2\pmod{3}$. Let $y=3k+2$, $k\in\mathbb Z$.
$$3x+3k+2=5702\iff 3(x+k)=5700$$ $$\iff x+k=1900\iff x=1900-k$$
All the integer solutions are $(x,y)=(1900-k,3k+2)$, $k\in\mathbb Z$.
$$x+y=(1900-k)+(3k+2)$$
$$=2k+1902\le 2003$$
$$\iff k\le 50.5\iff k\le 50$$
Natural numbers have different definitions among different authors (they do or don't include $0$; see Wikipedia or MathWorld).
Assuming $0\in\mathbb N$, you want to find all $k\in\mathbb Z$ such that $1900-k\ge 0$, $3k+2\ge 0$, $k\le 50$.
You'll get $k\in\{0,1,2,\ldots,50\}$, which will generate all the $(x,y)$ that the problem is asking about.
I.e., all the needed pairs are $$(x,y)\in\{(1900,2),(1899,5),(1898,8),\ldots,(1850,152)\}$$ There are $51$ such pair.
