6. 81 Function $f$ is defined on the real number domain, and satisfies the following conditions: for any real number $x$,
$$f(2+x)=f(2-x), f(7+x)=f(7-x) .$$

If $x=0$ is a root of $f(x)=0$, then $f(x)=0$ should have at least how many roots in the interval $-1000 \leqslant x \leqslant 1000$?
［Solution］From the given conditions, we have
$$\begin{array}{l}
f(x)=f(2+x-2)=f[2-(x-2)]=f(4-x), \\
f(4-x)=f[7-(x+3)]=f[7+(x+3)]=f(10+x)
\end{array}$$

From (1) and (2), we get $f(x)=f(10+x)$.
Thus, from $f(0)=0$, we have
$$f(0)=f( \pm 10)=f( \pm 20)=\cdots=f( \pm 1000)=0$$

This gives us 201 roots of $f(x)=0$ in the interval $[-1000,1000]$.
From (1), we also get $f(4)=f(0)=0$.
Then, from (3), we have
$$\begin{aligned}
f(4) & =f(4 \pm 10)=f(4 \pm 20) \\
& =\cdots=f(4 \pm 990) \\
& =f(4-1000)=0
\end{aligned}$$

This gives us another 200 roots of $f(x)=0$ in the interval $[-1000,1000]$.
Therefore, $f(x)=0$ has at least 401 roots in the interval $[-1000,1000]$.