Example 12 (1995 National High School Mathematics League Question) As shown in Figure 5-11, in the rectangular coordinate plane, the number of integer points (i.e., points with both coordinates as integers) that satisfy the inequality system $\left\{\begin{array}{l}y \leqslant 3 x, \\ y \geqslant \frac{1}{3} x, \\ x+y \leqslant 100\end{array}\right.$ is . $\qquad$
Solution: Fill in 2551. Reason: The geometric meaning of the system of inequalities $\left\{\begin{array}{l}y \leqslant 3 x, \\ y \geqslant \frac{1}{3} x, \quad \text { is: a triangular region } Q \text { formed by the three lines } \\ x+y \leqslant 100\end{array}\right.$ $y=3 x, y=\frac{1}{3} x$ and $x+y=100$. Clearly, region $Q$ is entirely within the first quadrant. To find the number of integer points within region $Q$, we can first consider the number of integer points within the region $\mathbf{R}$ formed by the $x$-axis, $y$-axis, and the line $x+y=100$, and then, based on symmetry, subtract twice the number of integer points within the region $G$ formed by the lines $y=\frac{1}{3} x, x+y=100$, and the $x$-axis, to find the required number of integer points.

Thus, the number of integer points (including the boundaries) within the region $\mathbf{R}$ formed by the $x$-axis, $y$-axis, and the line $x+y=100$ is $1+2+3+\cdots+100+101=\frac{(1+101) \times 101}{2}=5151$.

The number of integer points (excluding the integer points on the boundary $y=\frac{1}{3} x$) within the region $G$ formed by the lines $y=\frac{1}{3} x, x+y=100$, and the $x$-axis is $100+96+92+\cdots+8+4=1300$.

By symmetry, the region formed by the lines $y=3 x, x+y=100$, and the $y$-axis (excluding the integer points on the boundary $y=3 x$) also contains 1300 integer points.
Therefore, the number of integer points that satisfy the conditions is $5151-2 \cdot 1300=2551$.