Example 4 asks how many sets of positive integer solutions the equation $x+y+z=1981$ has?

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Solution 1 Since 1981 is an odd number, at least one of $x, y, z$ must be odd. Let $z$ be the odd number, then $z+y \leqslant 1980$. Therefore, the solutions to the original equation are the lattice points in the Cartesian coordinate system that satisfy $x+y \leqslant 1980$. Since $x, y$ are positive integers, these lattice points form a triangular array with 1981-2 points as the legs. Hence,
the total number of lattice points $=\frac{1+(1981-2)}{2}(1981-2)=1959210$.
From this, we know that the original equation has 1959210 sets of positive integer solutions.
Solution 2 Transform the original equation and let
$$
x+y=1981-z=k+1 .
$$

Since
$$
x \geqslant 1, y \geqslant 1 \text {, }
$$

then
$$
x+y \geqslant 2 \text {. }
$$
$$
z \geqslant 1 \text {, }
$$

Since
$$
1981-z \leqslant 1980 \text {. }
$$

Therefore, from (5), (7), and (8) we get
$$
1 \leqslant k \leqsl979 \quad(k \in \mathbf{N}) .
$$

From (5) we get
$$
y=k+1-x \text {. }
$$

From (6) we know
$$
1 \leqslant x \leqslant k \text {. }
$$

From (9) and (10) we get the number of positive integer solution sets for the original equation $x+y+z=1981$:
$$
p(k)=\sum_{k=1}^{1979} k=\frac{1979 \cdot 1980}{2}=1959210 .
$$