11. Given a sequence $\left\{a_{n}\right\}$ whose terms are all non-negative, the sum of its first $n$ terms is $S_{n}$, and for any $n \in N^{*}$, we have
$$
a_{n+1} \leq \frac{a_{n}+a_{n+2}}{2} \text {. }
$$
(1) If $a_{1}=1, a_{505}=2017$, find the maximum value of $a_{6}$;
(2) If for any $n \in N^{*}$, we have $S_{n} \leq 1$, prove that $0 \leq a_{n}-a_{n+1} \leq \frac{2}{n(n+1)}$.
11. (1) From the problem, we know that $a_{n+1}-a_{n} \leq a_{n+2}-a_{n+1}$. Let $d_{i}=a_{i+1}-a_{i}, 1 \leq i \leq 504$,

then $d_{1} \leq d_{1} \leq \ldots \leq d_{504}, d_{1}+d_{1}+\ldots+d_{504}=2016$, thus
$$
\frac{d_{1}+d_{2}+\ldots+d_{5}}{5} \leq \frac{d_{6}+d_{7}+\ldots+d_{504}}{499}=\frac{2016-\left(d_{1}+d_{2}+\ldots+d_{5}\right)}{499}
$$

Therefore, $d_{1}+d_{2}+\ldots+d_{5} \leq 20$, so $a_{1}=a_{1}+d_{1}+d_{2}+\ldots+d_{5} \leq 21$, equality holds for an arithmetic sequence.
(2) If there exists $k \in N_{+}$, such that $a_{k}b_{k+1}$ and $b_{k}>0$. Hence
$1 \geq a_{1}+\ldots+a_{n}=b_{1}+2 b_{2}+\ldots+n b_{n} \geq(1+2+\ldots+n) b_{n}=\frac{n(n+1)}{2} b_{n}$, thus $b_{n} \leq \frac{2}{n(n+1)}$, meaning for any positive integer $n$, we have $0 \leq a_{n}-a_{n+1} \leq \frac{2}{n(n+1)}$.