jmc

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 $$\{\begin{array}{ll}{r}_A+r_B& =26\\ b_A+b_B& =26\\ b_A& =6\cdot r_A\\ r_B& =m\cdot b_B\end{array}$$ 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 $$\{\begin{array}{ll}{r}_A+m\cdot b_B& =26\\ 6\cdot r_A+b_B& =26.\end{array}$$ Multiplying the first equation by 6 and subtracting, we get $$(6m-1)b_B=5\cdot 26=2\cdot 5\cdot 13.$$ 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 $r_B=m\cdot b_B=11\cdot 2=\boxed{22}$ red cards in pile $B.$