1. Arrange the numbers $1,2, \cdots, 13$ in a row $a_{1}, a_{2}$, $\cdots, a_{13}$, where $a_{1}=13, a_{2}=1$, and ensure that $a_{1}+a_{2}+$ $\cdots+a_{k}$ is divisible by $a_{k+1}(k=1,2, \cdots, 12)$. Then the value of $a_{4}$ $+a_{5}+\cdots+a_{12}$ is $\qquad$ .
II. 1.68.
Since $a_{1}+a_{2}+\cdots+a_{12}$ is divisible by $a_{13}$, then $a_{1}+a_{2}+\cdots+a_{12}+a_{13}$ is also divisible by $a_{13}$, meaning $a_{13}$ is a factor of $a_{1}+a_{2}+\cdots+a_{12}+a_{13}=13 \times 7$. Given that $a_{1}=13$, $a_{2}=1$, we have $a_{13}=7$.

Also, since $a_{1}+a_{2}$ is divisible by $a_{3}$, then $a_{3}$ is a factor of $a_{1}+a_{2}=2 \times 7$. Therefore, $a_{3}=2$. We can sequentially obtain a series of numbers that satisfy the conditions: $13,1,2,8,3,9,4,10,5,11,6,12,7$. Thus,
$$
a_{4}+a_{5}+\cdots+a_{12}=13 \times 7-13-1-2-7=68 .
$$