Problem 982

Let $(x, y, z)$ be a positive integer solution to (1), then $3 z \leqslant 2009$, i.e., $1 \leqslant z \leqslant 669$, which respectively yield
$$x+2 y=2009,2006, \cdots, 5$$

Correspondingly, the range of values for $y$ are
$$\begin{array}{l} 1 \leqslant y \leqslant 1004,1 \leqslant y \leqslant 1002,1 \leqslant y \leqslant 1001 \\ 1 \leqslant y \leqslant 999, \cdots, 1 \leqslant y \leqslant 2 \end{array}$$

Since when $y$ and $z$ are determined, the value of $x$ is uniquely determined, the number of positive integer solutions to (1) is
$$\begin{aligned} & (1004+1002)+(1001+999)+\cdots+(5+3)+2 \\ = & 2006+2000+\cdots+8+2 \\ = & \frac{1}{2}(2006+2) \times 335=336340 \end{aligned}$$

In summary, there are 336340 sets of positive integer solutions.