Let pile $ A $ have $ r_A $ red cards and $ b_A $ black cards, and let pile $ B $ have $ r_B $ red cards and $ b_B $ black cards. From the given information, we have $$\left\{ \begin{array}{ll} r_A+r_B & = 26 \\ b_A+b_B & = 26 \\ b_A &= 6\cdot r_A \\ r_B &= m\cdot b_B \\ \end{array} \right.$$ for some positive integer $ m $. Substituting $ 6\cdot r_A $ and $ m\cdot b_B $ for $ b_A $ and $ r_B, $ respectively, in the first two equations, we have $$\left\{ \begin{array}{ll} r_A+m\cdot b_B & = 26 \\ 6\cdot r_A+b_B & = 26.\end{array} \right.$$ Multiplying the first equation by 6 and subtracting, we get $$(6m-1)b_B=5\cdot26=2\cdot5\cdot13.$$ Since $ m $ is an integer, we have two possibilities: $ b_B=2 $ and $ m=11, $ or $ b_B=26 $ and $ m=1 $. The latter implies that pile $ A $ is empty, which is contrary to the statement of the problem, so we conclude that $ b_B=2 $ and $ m=11 $. Then, there are $ r_B=m\cdot b_B=11\cdot2=\boxed{22} $ red cards in pile $ B $.