(14) Let $x_{i} \in\{\sqrt{2}-1, \sqrt{2}+1\}, i=1,2, \cdots, 2010$. Let $S=x_{1} x_{2}+$ $x_{3} x_{4}+\cdots+x_{2009} x_{2010}$.
(1) Can $S$ equal 2010? Prove your conclusion;
(2) How many different integer values can $S$ take?
14 (1) Since
$$
\begin{array}{l}
(\sqrt{2}-1)^{2}=3-2 \sqrt{2}, \\
(\sqrt{2}+1)^{2}=3+2 \sqrt{2}, \\
(\sqrt{2}-1)(\sqrt{2}+1)=1,
\end{array}
$$

Therefore,
$$
x_{2 i-1} x_{2 i} \in\{3-2 \sqrt{2}, 3+2 \sqrt{2}, 1\} .
$$

Let the sum $S$ contain $a$ terms of $3+2 \sqrt{2}$, $b$ terms of $3-2 \sqrt{2}$, and $c$ terms of 1, where $a, b, c$ are non-negative integers, and $a+b+c=1005$.
$$
\begin{aligned}
S & =\sum_{i=1}^{1005} x_{2 i-1} x_{2 i} \\
& =(3+2 \sqrt{2}) a+(3-2 \sqrt{2}) b+c \\
& =3 a+3 b+c+2 \sqrt{2}(a-b) .
\end{aligned}
$$

If $S=2010$, then $a=b$, at this time
$$
S=6 a+c=6 a+(1005-a-a)=4 a+1005
$$

is an odd number. Therefore, $S$ cannot be equal to 2010.
(2) From (1), we know that if $S$ is an integer, then $a=b, S=4 a+1005$. Since $a+b+c=2 a+c=1005, 0 \leqslant a \leqslant 502$, $S$ can take 503 different integer values.