Three. (25 points) Let the quadratic function
$$
f(x)=a x^{2}+b x+c
$$

satisfy the following conditions:
(i) When $x$ is a real number, its minimum value is 0, and
$$
f(x-1)=f(-x-1)
$$

holds;
(ii) There exists a real number $m (m>1)$, such that there exists a real number $t$, as long as $1 \leqslant x \leqslant m$, then $f(x+t) \leqslant x$ holds.
(1) In condition (ii), let $x=1$, we get $f(1)=1$.
From condition (i), we know that the quadratic function opens upwards and is symmetric about $x=-1$, so we can assume the quadratic function to be
$$
f(x)=a(x+1)^{2}(a>0) .
$$

Substituting $f(1)=1$ into the above equation, we get $a=\frac{1}{4}$.
Therefore, $f(x)=\frac{1}{4}(x+1)^{2}$.
(2) Suppose there exists $t \in \mathbf{R}$, such that for $1 \leqslant x \leqslant m$, we have $f(x+t) \leqslant x$.
Taking $x=1$, we have
$$
(t+2)^{2} \leqslant 4 \Rightarrow-4 \leqslant t \leqslant 0 \text {. }
$$

Taking $x=m$, we have
$$
\begin{array}{l}
(m+t+1)^{2} \leqslant 4 m \\
\Rightarrow m^{2}+2(t-1) m+\left(t^{2}+2 t+1\right) \leqslant 0 \\
\Rightarrow 1-t-\sqrt{-4 t} \leqslant m \leqslant 1-t+\sqrt{-4 t} .
\end{array}
$$

Thus, $m \leqslant 1-t+\sqrt{-4 t}$
$$
\leqslant 1-(-4)+\sqrt{-4 \times(-4)}=9 \text {. }
$$

When $t=-4$, for any $x(1 \leqslant x \leqslant 9)$, we always have
$$
\begin{array}{l}
f(x-4)-x=\frac{1}{4}\left(x^{2}-10 x+9\right) . \\
=\frac{1}{4}(x-1)(x-9) \leqslant 0 .
\end{array}
$$

Therefore, the maximum value of $m$ is 9.
(Sun Yan, Anhui Provincial Anqing City Educational Research Office, 246003, Huang Quanfu, Anhui Provincial Huaining County Jiangzhen Middle School, 246142)