Example 7 (2006 Turkey National Team Selection Test) Given the integer sequence $\left\{x_{n}\right\}$ satisfies
$$
x_{n+1}=x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}, n \geqslant 1 .
$$

Find the minimum value of $x_{1}$ such that 2006 divides $x_{2006}$.
Notice that
$x_{n+1}=\sum_{i=1}^{n} x_{n}^{2}=x_{n}+x_{n}^{2}=x_{n}\left(x_{n}+1\right)$.
Therefore, $2 \mid x_{n+1}$.
Hence, $2006\left|x_{2006} \Leftrightarrow 1003\right| x_{2006} \Leftrightarrow 17 \mid x_{2006}$ and $59 \mid x_{2006}$.
If $x_{n} \equiv c(\bmod 17)$, then
$x_{n}\left(x_{n}+1\right) \equiv 0(\bmod 17) \Leftrightarrow c \equiv 0(\bmod 17)$ or $c \equiv-1(\bmod 17)$.
However, $x(x+1) \equiv-1(\bmod 17)$ has no solution, so
if $17 \mid x_{2006}$, then
$x_{2006} \equiv x_{2005}\left(x_{2005}+1\right)(\bmod 17)$.
If $x_{2005} \equiv-1(\bmod 17)$, then $x_{2004}$ has no solution.
Therefore, $17 \mid x_{2005}$.
By the same reasoning, $17 \mid x_{2}$, thus
$x_{1} \equiv 0(\bmod 17)$ or $x_{1} \equiv-1(\bmod 17)$.
If $x_{n} \equiv c(\bmod 59)$, then
$x_{n}\left(x_{n}+1\right) \equiv 0(\bmod 59) \Leftrightarrow c \equiv 0(\bmod 59)$ or $c \equiv-1(\bmod 59)$.
However, $(x+1) x \equiv-1(\bmod 59)$ has no solution, so similarly,
$x_{1} \equiv 0$ or $-1(\bmod 59)$.
If $x_{1} \equiv 0(\bmod 59), x_{1} \equiv 0(\bmod 17)$, then, the minimum value of $x_{1}$ is 1003;
If $x_{1} \equiv-1(\bmod 59), x_{1} \equiv 0(\bmod 17)$, then, the minimum value of $x_{1}$ is $15 \times 59-1=884$;
If $x_{1} \equiv-1(\bmod 17), x_{1} \equiv 0(\bmod 59)$, then, the minimum value of $x_{1}$ is 118;
If $x_{1} \equiv-1(\bmod 17), x_{1} \equiv-1(\bmod 59)$, then, the minimum value of $x_{1}$ is 1002.
Therefore, the minimum value of $x_{1}$ is 118.