Example 6 Function $f$ 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 $f(7-x)=f(7+x)$. Suppose $x=0$ is a root of $f(x)=0$, and let $N$ be the number of roots of $f(x)=0$ in the interval $[-1000,1000]$. Find the minimum value of $N$.
From the problem, we know that the graph of the function $f(x)$ is symmetric about the lines $x=2$ and $x=7$, so $f(4) = f(2+2) = f(2-2) = f(0) = 0$, and $f(10) = f(7+3) = f(7-3) = f(4) = 0$. Therefore, $f(x) = 0$ has at least two roots in the interval $(0,10]$. On the other hand, from $\left\{\begin{array}{l}f(2-x) = f(2+x) \\ f(7-x) = f(7+x)\end{array}\right.$, we can derive $\left\{\begin{array}{l}f(x) = f(4-x) \\ f(x) = f(14-x)\end{array}\right.$, so $f(4-x) = f(14-x)$, which means $f(x) = f(x+10)$. Thus, the function $y = f(x)$ is a periodic function with a period of $T = 10$, and therefore $f(x) = 0$ has at least $200 \times 2 + 1 = 401$ roots in the interval $[-1000, 1000]$. As shown in Figure 6-6, we can construct a "sawtooth" function $y = f(x)$ that satisfies all the above conditions, and the equation $f(x) = 0$ has 401 roots in the interval $[-1000, 1000]$, with no other roots. Therefore, the minimum value of $N$ is 401.