13. In the sequence $\left\{a_{n}\right\}$, $a_{n}=a \cdot 2^{n}+b \cdot n-80$, where $a, b \in \mathbf{N}_{+}$. It is known that the sum of the first $n$ terms $S_{n}$ reaches its minimum value if and only if $n=6$, and $a_{36}$ is divisible by 7. Find the value of $\left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{12}\right|$.
$$
\left\{\begin{array} { l } 
{ 2 ^ { 6 } a + 6 b - 8 0 0 , } \\
{ 7 | 2 ^ { 3 6 } a + 3 6 b - 8 0 }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ 6 4 a + 6 b - 8 0 0 , } \\
{ a + b = 7 k + 3 }
\end{array} \Rightarrow \left\{\begin{array}{l}
a=1, \\
b=2 .
\end{array}\right.\right.\right.
$$

Thus, \( a_{n}=2^{n}+2 n-80 \). Therefore, \( \left|a_{1}\right|+\left|a_{2}\right|+\cdots+\left|a_{12}\right|=S_{12}-2 S_{6} \)
$$
\begin{array}{l}
=2^{13}-2+12 \times 13-12 \times 80-2 \cdot\left(2^{7}-2+6 \times 7-6 \times 80\right) \\
=2^{13}-2^{8}+2+12 \times 6=8010 .
\end{array}
$$