Example 7 The function $f(x)$ is defined on the set of real numbers, and for all real numbers $x$ it satisfies the equations: $f(2+x)=f(2-x)$ and $f(x+7)=f(7-x)$. Suppose $x=0$ is a root of $f(x)=0$, and let $N$ denote the number of roots of $f(x)=0$ in the interval $[-1000,1000]$. Find the minimum value of $N$.
$$
\begin{array}{l}
\text { Solution: } \because f(x+2)=f(2-x), \\
\begin{array}{l}
\therefore f(x+4)=f[(x+2)+2] \\
\quad=f[2-(x+2)]=f(-x) . \\
\because f(x+7)=f(7-x), \\
\therefore f(x+14)=f[(x+7)+7]
\end{array} \\
\quad=f[7-(x+7)]=f(-x) .
\end{array}
$$

Thus, $f(x+14)=f(4+x)$,
which means $f(x+10)=f(x)$.
Therefore, $f(x)$ is a periodic function with a period of 10.
$\because x=0$ is a root of $f(x)=0$, then $x= \pm 10$, $\pm 20, \cdots, \pm 1000$ are all roots of $f(x)=0$, totaling 201.

Also, $\because f(4)=f(0+4)=f(-0)=0$, i.e., $x$ $=4$ is a root of $f(x)=0$, then $x=4 \pm 10, \cdots$, $4 \pm 10 k(-1000 \leqslant 4 \pm 10 k \leqslant 1000)$ are all roots of $f(x)$ $=0$, totaling 200.
Therefore, the minimum value of $N$ is 401.
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