Three. (25 points) Let $n$ be an integer. If there exist integers $x, y, z$ satisfying
$$
n=x^{3}+y^{3}+z^{3}-3 x y z \text {, }
$$

then $n$ is said to have property $P$.
(1) Determine whether $1, 2, 3$ have property $P$;
(2) Among the 2014 consecutive integers $1, 2, \cdots, 2014$, how many do not have property $P$?
(1) Let $x=1, y=z=0$, we get
$$
1=1^{3}+0^{3}+0^{3}-3 \times 1 \times 0 \times 0 \text {. }
$$

Thus, 1 has property $P$.
Let $x=y=1, z=0$, we get
$$
2=1^{3}+1^{3}+0^{3}-3 \times 1 \times 1 \times 0 \text {. }
$$

Thus, 2 has property $P$.
If 3 has property $P$, then there exist integers $x, y, z$ such that
$$
\begin{array}{l}
3=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx) \\
\Rightarrow 3\left|(x+y+z)^{3} \Rightarrow 3\right|(x+y+z) \\
\Rightarrow 9 \mid\left[(x+y+z)^{3}-3(x+y+z)(xy+yz+zx)\right] \\
\Rightarrow 9 \mid 3,
\end{array}
$$

which is impossible.
Therefore, 3 does not have property $P$.
(2) Let $f(x, y, z)=x^{3}+y^{3}+z^{3}-3xyz$. Then
$$
\begin{array}{l} 
f(x, y, z)=(x+y)^{3}+z^{3}-3xy(x+y)-3xyz \\
=(x+y+z)^{3}-3z(x+y)(x+y+z)- \\
3xy(x+y+z) \\
=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx) \\
= \frac{1}{2}(x+y+z)\left(x^{2}+y^{2}+z^{2}-xy-yz-zx\right) \\
= \frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right],
\end{array}
$$

i.e.,
$$
\begin{array}{l}
f(x, y, z) \\
=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] .
\end{array}
$$

Assume without loss of generality that $x \geqslant y \geqslant z$.
If $x-y=1, y-z=0, x-z=1$, i.e.,
$$
x=z+1, y=z \text {, }
$$

then $f(x, y, z)=3z+1$;
If $x-y=0, y-z=1, x-z=1$, i.e.,
$$
x=y=z+1 \text {, }
$$

then $f(x, y, z)=3z+2$;
If $x-y=1, y-z=1, x-z=2$, i.e., $x=z+2, y=z+1$,
then $f(x, y, z)=9(z+1)$.
Thus, numbers of the form $3k+1$ or $3k+2$ or $9k$ $(k \in \mathbf{Z})$ all have property $P$.
Note that,
$$
\begin{array}{l}
f(x, y, z) \\
=(x+y+z)^{3}-3(x+y+z)(xy+yz+zx) .
\end{array}
$$

If $3 \mid f(x, y, z)$, then
$3\left|(x+y+z)^{3} \Rightarrow 3\right|(x+y+z)$
$\Rightarrow 9 \mid f(x, y, z)$.
In summary, integers $n$ do not have property $P$ if and only if $n=9k+3$ or $n=9k+6$ $(k \in \mathbf{Z})$.

Since $2014=9 \times 223+7$, in the 2014 consecutive integers from 1 to 2014, the number of integers that do not have property $P$ is $224 \times 2=448$.