jmc

In Sally's arrangement, the number of candies is $ab+2a+b$. In Rita's arrangement, the number of candies is $\left(5a-4\right)\left(\frac{b-1}{3}\right)$. The number of candies didn't change, so these two expressions are equal. Therefore, $$\begin{array}{rl}ab+2a+b& =(5a-4)\left(\frac{b-1}{3}\right)\Rightarrow \\ 3ab+6a+3b& =(5a-4)(b-1)\Rightarrow \\ 3ab+6a+3b& =5ab-4b-5a+4\Rightarrow \\ 0& =2ab-7b-11a+4\Rightarrow \\ -4& =b(2a-7)-11a\Rightarrow \\ -4+\frac{11}{2}(7)& =b(2a-7)-\frac{11}{2}(2a-7)\Rightarrow \\ \frac{-8}{2}+\frac{77}{2}& =\left(b-\frac{11}{2}\right)(2a-7)\Rightarrow \\ 69& =(2b-11)(2a-7).\end{array}$$The prime factorization of $69$ is $3\cdot 23$. So we have the following possibilities. \begin{array}{c|c|c|c|c|c} $2a-7%%DISP_1%%amp;$2b-11%%DISP_1%%amp;$2a%%DISP_1%%amp;$2b%%DISP_1%%amp;$a%%DISP_1%%amp;$b$\\ \hline $1%%DISP_1%%amp;$69%%DISP_1%%amp;$8%%DISP_1%%amp;$80%%DISP_1%%amp;$4%%DISP_1%%amp;$40$\\ $3%%DISP_1%%amp;$23%%DISP_1%%amp;$10%%DISP_1%%amp;$34%%DISP_1%%amp;$5%%DISP_1%%amp;$17$\\ $23%%DISP_1%%amp;$3%%DISP_1%%amp;$30%%DISP_1%%amp;$14%%DISP_1%%amp;$15%%DISP_1%%amp;$7$\\ $69%%DISP_1%%amp;$1%%DISP_1%%amp;$76%%DISP_1%%amp;$12%%DISP_1%%amp;$38%%DISP_1%%amp;$6$ \end{array}We know from above, since Rita's arrangement must have integral dimensions, that $b-1$ is divisible by $3$. A check shows that the pairs $(a,b)$ that don't work are $(5,17)$ and $(38,6)$. Thus we have either $(a,b)=(15,7)$ or $(a,b)=(4,40)$. There are $ab+2a+b$ candies. In this first case we have $(15)(7)+2(15)+7=142$ candies. In the second case there are $(4)(40)+2(4)+40=208$ candies. Thus the maximum number of candies that could be in Sally's bag is $\boxed{208}$.