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We have $3x+5y=1008 \Leftrightarrow 3(x-1)+5y=1005$ .

Since $5x\equiv1005\equiv0\pmod{5}$ , we get $3(x-1)\equiv0\pmod{5}$ or $x\equiv 1\pmod{5}$ .

Let $x=5k+1$ , where $k$ is an integer.

Thus, $y=\dfrac{1005-3(x-1)}{5}=\dfrac{1005-15k}{5}=201-3k$ .

Because $x,y>0$ , we have: $\left\{\begin{array}{l}5k+1>0\\201-3k>0\end{array}\right. \Leftrightarrow 0\le x\le 66$ .

So, the equation has $\boxed{67}$ positive integer solutions.

$x=336-\frac{5y}{3}\Rightarrow 3|y$ .

Also $x,y>0$ , so, $0<336-\frac{5y}{3} \Rightarrow \frac{5y}{3} <336 \Rightarrow 0<y<201$ .

The number of solutions is $\frac{201}{3}=\boxed{67}$ which includes all $y$ divisible by $3$ , greatest than zero but less than 201.

Solving for $y$ in terms $x$ produces: $y = \frac{1008 - 3x}{5} = \frac{3(336 - x)}{5}.$ In order for $y$ to be a positive integer, we must have $5|(336 - x)$ . This yields $x = {1, 6, 11, 16,..., 326, 331} \Rightarrow x = 5k - 4; k = 1, 2,..., 67.$ Thus there are $\boxed{67}$ ordered-pairs $(x, y); x, y \in \mathbb{N}.$