11. Let the sequence of positive integers $\left\{a_{n}\right\}$ have the following properties simultaneously:
(i) For any positive integer $k$, we have $a_{2 k-1}+a_{2 k}=2^{k}$;
(ii) For any positive integer $m$, there exists a positive integer $l \leqslant m$, such that $a_{m+1}=\sum_{i=l}^{m} a_{i}$. Find the maximum value of $a_{2}+a_{4}+a_{6}+\cdots+a_{2022}$.
For $k=1$, $a_{1}+a_{2}=2 \Rightarrow a_{1}=a_{2}=1$; for $k=2$, $a_{3}+a_{4}=4$. According to the problem, we have $a_{m+1} \geqslant a_{m}$, so $a_{3}=1, a_{4}=3$ or $a_{3}=a_{4}=2$. Conjecture that for $k \geqslant 2$, $a_{2 k}=2^{k-1}$ or $3 \times 2^{k-2}$.

Assume the conclusion holds for $k (k \geqslant 2)$, and consider the case for $k+1$.
By (ii), there exists a positive integer $l \leqslant 2 k$ such that $a_{2 k+1}=\sum_{i=l}^{2 k} a_{i}$. If $l \leqslant 2 k-1$, then $2^{k}=a_{2 k-1}+a_{2 k} \leqslant a_{2 k+1} \leqslant \frac{a_{2 k+1}+a_{2 k+2}}{2}=2^{k}$, so all the equalities hold, i.e., $a_{2 k+1}=a_{2 k-1}+a_{2 k}=2^{k}$.
Thus, $a_{2 k+2}=2^{k+1}-a_{2 k+1}=2^{k}$, and the conclusion holds; if $l=2 k$, then $a_{2 k+1}=a_{2 k} \Rightarrow a_{2 k+2}=2^{k+1}-a_{2 k}$.
When $a_{2 k}=2^{k-1}$, $a_{2 k+2}=2^{k+1}-2^{k-1}=3 \times 2^{k-1}$, and the conclusion holds; when $a_{2 k}=3 \times 2^{k-2}$, $a_{2 k+2}=2^{k+1}-3 \times 2^{k-2}=5 \times 2^{k-2}$,
at this point, $a_{2 k+1}=a_{2 k}=3 \times 2^{k-2}, a_{2 k+1}<a_{2 k+2}<a_{2 k}+a_{2 k+1}$, so there does not exist $l_{1} \leqslant 2 k+1$ such that $a_{2 k+2}=\sum_{i=l_{1}}^{2 k+1} a_{i}$, which is a contradiction. Thus, the conclusion also holds for $k+1$.
By the principle of mathematical induction, for $k \geqslant 2$, $a_{2 k}=2^{k-1}$ or $3 \times 2^{k-2}$.
Therefore, $a_{4 k}=2^{2 k-1}$ or $3 \times 2^{2 k-2}, a_{4 k+2}=2^{2 k}$ or $3 \times 2^{2 k-1}$, and from the proof process, when $a_{4 k}=3 \times 2^{2 k-2}$, $a_{4 k+2} \neq 3 \times 2^{2 k-1}$, then
$$
\begin{array}{l}
a_{4 k}+a_{4 k+2} \leqslant \max \left\{2^{2 k-1}+2^{2 k}, 2^{2 k-1}+3 \times 2^{2 k-1}, 3 \times 2^{2 k-2}+2^{2 k}\right\} \\
=2^{2 k-1}+3 \times 2^{2 k-1}=2^{2 k+1} .
\end{array}
$$

Thus, $a_{2}+a_{4}+a_{6}+\cdots+a_{2022}=a_{2}+\sum_{i=1}^{505}\left(a_{4 i}+a_{4 i+2}\right) \leqslant 1+\sum_{i=1}^{505} 2^{2 i+1}$
$=1+\frac{2^{3}\left(4^{505}-1\right)}{4-1}=\frac{2^{1013}-5}{3}$, equality holds when $a_{4 i}=2^{2 i-1}, a_{4 i+2}=3 \times 2^{2 i-1}$
$\Rightarrow a_{4 i+1}=2^{2 i-1}, a_{4 i-1}=2^{2 i-1}(i=1,2, \cdots, 505), a_{1}=a_{2}=1$.
Therefore, the maximum value of $a_{2}+a_{4}+a_{6}+\cdots+a_{2022}$ is $\frac{2^{1013}-5}{3}$.