Solve: First, find a particular solution of (1).
$$x=\frac{1}{7}(2012-19 y)=287-3 y+\frac{1}{7}(3+2 y) .$$

Therefore, $\frac{1}{7}(3+2 y)$ must be an integer. Taking $y_{0}=2$, then $x_{0}=282$.
Using the conclusion of Theorem 2, the general solution of equation (1) is
$$\left\{\begin{array}{l} x=282-19 t, \\ y=2+7 t . \end{array} \text { ( } t \text { is an integer }\right)$$

Combining $x>0, y>0$ and $t$ being an integer, we can solve to get $0 \leqslant t \leqslant 14$.
Therefore, equation (1) has 15 sets of positive integer solutions.