Let the first 4 digit number have digits a, b, c and d in order.
This four digit number can be expressed as 1000a + 100b + 10c + d.
Then, when you move the first digit to the end, and move the other digits up by one, the digits are b, c, d and a in order.
This four digit number can be expressed as 1000b +100c + 10d + a.
When you add these two four digit numbers together, you get 1000a + 100b + 10c + d + 1000b + 100c + 10d + a, which simplified gives 1001a + 1100b + 110c + 11d.
From this we can factorise out an 11, as all of 1001, 1100, 110 and 11 go into 11, factoring to become 11(91a + 100b + 10c + d).
Thus, as 11 multiplied by an integer is a multiple of 11, the sum of any 4 digit number and it's single permutation is always a multiple of 11. QED!!!