11. (20 points) Given the set of integers
$$
A=\{1,2, \cdots, 100\} \text {. }
$$

Let the function $f: A \rightarrow A$ satisfy:
(1) For any $1 \leqslant i \leqslant 99$, we have
$$
|f(i)-f(i+1)| \leqslant 1 \text {; }
$$
(2) For any $1 \leqslant i \leqslant 100$, we have
$$
\begin{array}{l}
f(f(i))=100 . \\
\text { Find the minimum possible value of } \sum_{i=1}^{100} f(i) \text { . }
\end{array}
$$
11. Let the set $B$ be the range of $f(x)$.

From (2), we know $B \neq \varnothing$.
Without loss of generality, assume
$$
B=\left\{a_{1}, a_{2}, \cdots, a_{t}\right\}\left(a_{1}<a_{2}<\cdots<a_{t}, t \geqslant 1\right) .
$$

From $f(f(i))=100 \Rightarrow 100 \in B$.
From (1), we know that the $t$ numbers in $B$ must be $t$ consecutive numbers. Assume $B=\{100,99, \cdots, 101-t\}$.
From (2), we know that for any $i \in B, f(i)=100$.
Then $f(100-t) \geqslant 99, f(99-t) \geqslant 98$,
......
Thus, $\sum_{i=1}^{100} f(i)$
$\geqslant 100 t+(99+98+\cdots+(101-t))+$ $(101-2 t)(101-t)$
$=100 t+\frac{(200-t)(t-1)}{2}+(101-2 t)(101-t)$
$=\frac{3}{2} t^{2}-\frac{205}{2} t+10101$
$\geqslant \frac{3}{2} \times 34^{2}-\frac{205}{2} \times 34+10101$
$$
=8350 \text {, }
$$

When $t=34$, $B=\{100,99, \cdots, 67\}$, at this time,
$$
\begin{array}{l}
f(i)=100(i=67,68, \cdots, 100), \\
f(66)=99, \cdots, f(34)=67, \\
f(i)=67(i=1,2, \cdots, 33),
\end{array}
$$

The minimum value sought is 8350.