Sandy has $384615 and Sue has $538461

Let X and Y be Sandy's and Sue's money. Let Z be the number of dollars Sandy's statement.

X+Z = Y-Z and 2(X-Z) = Y+Z
X = 5Z
Y = 7Z

Let A be the first digit in Sue's wealth, B be the remaining digits, and n be the number of digits

Y = 10^(n-1)A+B
X = 10B + A

7X = 5Y
7(10B+A) = 5(10^(n-1)A+B)
(5*10^(n-1)-7)A=65B

Since 65B is divisible by 5 and 5*10^(n-1)-7 is not, A must equal 5

5*10^(n-1)-7 = 13B

The least value of n such that 5*10^(n-1)-7 is a multiple of 13 is 6:

499993=38461*13

The next is 12, which makes them have more than 1 million dollars. Hence n=6, B=38461