16 In the sequence $\left\{a_{n}\right\}$, $a_{1}, a_{2}$ are given non-zero integers, $a_{n+2}=\left|a_{n+1}-a_{n}\right|$.
(1) Let $a_{1}=2, a_{2}=-1$, find $a_{2008}$;
(2) Prove: From $\left\{a_{n}\right\}$, it is always possible to select infinitely many terms to form two different constant sequences.
(16) (1) From $a_{1}=2, a_{2}=-1, a_{3}=3, a_{4}=4, a_{5}=1, a_{6}=3, a_{7}=2$, $a_{8}=1, a_{9}=1, a_{10}=0, a_{11}=1, a_{12}=1, a_{13}=0, \cdots$, we can see that starting from the 8th term, every three consecutive terms cyclically take the values $1, 1, 0$, hence $a_{2008}=0$.
(2) First, we prove that the sequence $\left\{a_{n}\right\}$ must have a zero term after a finite number of terms. Assume that $\left\{a_{n}\right\}$ contains no zero terms. Since $a_{n+2}=\left|a_{n+1}-a_{n}\right|$, for $n \geqslant 3$, we have $a_{n} \geqslant 1$. When $a_{n+1}>a_{n}$,
$$
a_{n+2}=a_{n+1}-a_{n} \leqslant a_{n+1}-1(n \geqslant 3) ;
$$

When $a_{n+1}a_{2 n+2}\right), \\
a_{2 n+2}\left(a_{2 n+1}0$ contradiction, thus $\left\{a_{n}\right\}$ must contain a zero term.

If the first zero term is $a_{n}$, let $a_{n-1}=M(M \neq 0)$, then starting from the $n$-th term, every three consecutive terms cyclically take the values $0, M, M$, i.e.,
$$
\left\{\begin{array}{l}
a_{n+3 k}=0, \\
a_{n+3 k+1}=M,(k=0,1,2, \cdots) . \\
a_{n+3 k+2}=M
\end{array}\right.
$$

Therefore, the sequence $\left\{a_{n}\right\}$ can definitely select infinitely many terms to form two different constant sequences.