11. (20 points) Given $n=d_{1} d_{2} \cdots d_{2017}$, where,
$$
\begin{array}{l} 
d_{i} \in\{1,3,5,7,9\}(i=1,2, \cdots, 2017), \text { and } \\
\sum_{i=1}^{1009} d_{i} d_{i+1} \equiv 1(\bmod 4), \\
\sum_{i=1010}^{2016} d_{i} d_{i+1} \equiv 1(\bmod 4) .
\end{array}
$$

Find the number of $n$ that satisfy the conditions.
11. Let $d_{k+1}=d_{1}, d_{k}=d_{0}$.

Since $d_{i} \equiv 1$ or $-1(\bmod 4)$, when replacing -1 with 1, the difference in $\sum_{i=1}^{k} d_{i} d_{i+1}$ is
$$
\begin{array}{l}
\Delta d_{i} \cdot\left(d_{i-1}+d_{i+1}\right)(\bmod 4)(i=1,2, \cdots, k), \\
\Delta d_{i} \equiv \pm 2(\bmod 4), \\
d_{i-1}+d_{i+1} \equiv 0, \pm 2(\bmod 4),
\end{array}
$$

Thus, the value of $\sum_{i=1}^{k} d_{i} d_{i+1}$ modulo 4 remains unchanged, so we can assume $d_{i} \equiv 1(\bmod 4)$.
Then $\sum_{i=1}^{k} d_{i} d_{i+1} \equiv k(\bmod 4)$.
Therefore, $\sum_{i=1}^{1009} d_{i} d_{i+1} \equiv 1010-d_{1010} d_{1} \equiv 1(\bmod 4)$,
$$
\begin{array}{l}
\sum_{i=1010}^{2016} d_{i} d_{i+1} \\
\equiv 2018-d_{2018} d_{1}-\left(1011-d_{1010} d_{1}\right) \\
\equiv 2017-d_{2017} d_{1}-1 \equiv 1(\bmod 4) .
\end{array}
$$

Thus, $d_{2018} d_{1} \equiv-1(\bmod 4)$, except for $d_{1}, d_{1010},$ and $d_{2017}$, the other $d_{i}$ can be any of $1, 3, 5, 7, 9$.
If $d_{1} \equiv 1(\bmod 4)$, then
$$
d_{1010} \equiv 1(\bmod 4), d_{2017} \equiv-1(\bmod 4),
$$
$n$ has $3 \times 3 \times 2 \times 5^{2014}$ values;
If $d_{1} \equiv-1(\bmod 4)$, then
$$
d_{1010} \equiv-1(\bmod 4), d_{2017} \equiv 1(\bmod 4),
$$
$n$ has $2 \times 2 \times 3 \times 5^{2014}$ values.
In summary, the number of $n$ is $6 \times 5^{2015}$.