11. (20 points) The function $f$ defined on $\mathbf{R}$ satisfies
$$
f(1+x)=f(9-x)=f(9+x) \text {. }
$$

If $f(0)=0$, and $f(x)=0$ has $n$ roots in the interval $[-4020$, 4020 ], find the minimum value of $n$.
11. By substituting $4-x$ for $x$ in $f(1+x)=f(9-x)$, we get
$$
f(5-x)=f(5+x).
$$

Thus, $x=5$ is a line of symmetry for $f(x)$.
Similarly, $x=9$ is also a line of symmetry for $f(x)$.
$$
\begin{array}{l}
\text { Also, } f(x)=f(1+(x-1)) \\
=f(9+(x-1))=f(x+8),
\end{array}
$$

so 8 is a period of $f(x)$.
Since $f(0)=0$, we have
$$
\begin{array}{l}
f(8)=0, \\
f(2)=f(5-3)=f(5+3)=f(8)=0.
\end{array}
$$

Thus, $f(x)$ has at least two roots in the interval $[8k, 8(k+1)](k \in \mathbf{Z})$.

Since $4020=8 \times 502+4$, the equation $f(x)=0$ has at least $502 \times 2+2=1006$ roots in the interval $[0,4020]$, and at least $502 \times 2=1004$ roots in the interval $[-4020,0)$.

Therefore, $f(x)=0$ has at least $1006+1004=2010$ roots in the interval $[-4020,4020]$.

A "sawtooth" function (as shown in Figure 6) can be constructed to satisfy all the above conditions, having 2010 roots in the interval $[-4020,4020]$, and no other roots. Hence, the minimum value of $n$ is 2010.