Three. (25 points) Let $n$ be an integer. If there exist integers $x$, $y$, $z$ satisfying $n=x^{3}+y^{3}+z^{3}-3 x y z$, then $n$ is said to have property $P$.

Among the numbers 1, 5, 2013, 2014, which numbers have property $P$, and which numbers do not have property $P$? Explain your reasoning.
Three, taking $x=1, y=z=0$, we get
$$
1=1^{3}+0^{3}+0^{3}-3 \times 1 \times 0 \times 0 \text {. }
$$

Thus, 1 has property $P$.
Taking $x=y=2, z=1$, we get
$$
5=2^{3}+2^{3}+1^{3}-3 \times 2 \times 2 \times 1 \text {. }
$$

Therefore, 5 has property $P$.
Next, consider numbers with property $P$.
$$
\begin{array}{l} 
\text { Let } f(x, y, z)=x^{3}+y^{3}+z^{3}-3 x y z \text {. Then } \\
f(x, y, z)=(x+y)^{3}+z^{3}-3 x y(x+y)-3 x y z \\
=(x+y+z)^{3}-3 z(x+y)(x+y+z)- \\
3 x y(x+y+z) \\
=(x+y+z)^{3}-3(x+y+z)(x y+y z+z x) \\
= \frac{1}{2}(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \\
= \frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right],
\end{array}
$$

That is, $f(x, y, z)$
$$
=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \text {. }
$$

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 $f(x, y, z)=3 z+1$;
If $x-y=0, y-z=1, x-z=1$, i.e., $x=y=z+1$,
then $f(x, y, z)=3 z+2$;
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 $3 k+1$ or $3 k+2$ or $9 k$ $(k \in \mathbf{Z})$ all have property $P$.
Therefore, 1, 5, and 2014 all have property $P$.
If 2013 has property $P$, then there exist integers $x, y, z$ such that
$$
2013=(x+y+z)^{3}-3(x+y+z)(x y+y z+z x) \text {. }
$$

Note that, $3 \mid 2013$.
Then $3\left|(x+y+z)^{3} \Rightarrow 3\right|(x+y+z)$
$$
\Rightarrow 9 \mid\left[(x+y+z)^{3}-3(x+y+z)(x y+y z+z x)\right]
$$
$\Rightarrow 9 \mid 2013$.
But $2013=9 \times 223+6$, a contradiction.
Thus, 2013 does not have property $P$.