Example 9 Let the plane region $D$ be represented by $N(D)$, which denotes the number of all integer points (i.e., points on the xoy plane where both coordinates $x$ and $y$ are integers) belonging to $D$. If $A$ represents the region enclosed by the curve $y=x^{2} (x \geqslant 0)$ and the two lines $x=10$, $y=1$ (including the boundaries), and $B$ represents the region enclosed by the curve $y=x^{2} (x \geqslant 0)$ and the two lines $x=1$, $y=100$ (including the boundaries), then $N(A \cup B)+N(A \cap B)=$ $\qquad$
$(1992$, Shanghai Senior High School Mathematics Competition)
Solution: Draw the figure in the Cartesian coordinate system, and it is easy to calculate from the figure
$$
\begin{array}{l}
N(A)=1^{2}+2^{2}+\cdots+10^{2} \\
=\frac{1}{6} \times 10(10+1)(2 \times 10+1)=385, \\
N(B)=\left(101-1^{2}\right)+\left(101-2^{2}\right)+\cdots \\
+\left(101-10^{2}\right) \\
=101 \times 10-\left(1^{2}+2^{2}+\cdots+10^{2}\right) \\
=1010-385=625. \\
\end{array}
$$
$$
N(A \cap B)=10.
$$

By $N(A \cup B)=N(A)+N(B)-N(A \cap B)$, we get
$$
\begin{array}{l}
N(A \cup B)=385+625-10=1000. \\
\therefore N(A \cup B)+N(A \cap B)=1010.
\end{array}
$$