Any two digit integer is in the form of xy.
The value of xy is = 10x + y.
Now switch the two around, you get yx.
The value of yx is = 10y + x.
Now assume yx > xy.
Then yx - xy = (10y + x) - (10x + y) = 10y + x -10x -y = 9y - 9x = 9 (y-x)
Since we assumed that yx > xy, then we see that 9 (y-x) is positive and also divisible by 9.
If xy > yx, it's the exact same calculation, but you get 9(x-y) instead.
EDIT: I thought you were asking only for the 2 digit case. Apparently you are asking for the n digit case. The proof for n digits is more complicated but similar.
Assume you have an n digit number x1 x2 ..... xn.
Its value is (x1 . 10^n) + (x2 . 10^(n-1)) + ..... ( xn . 10^0)
And its reverse has value (xn . 10^n) + (xn-1 . 10^(n-1)) + ..... (x1 . 10^0)
When you subtract one from another, you get a long number, where each number has a coefficient and a subtraction of two powers of 10, which is a power of 9.
In order to conclude the proof, you need to prove the lemma that 10^(n-k) - 10^(k) | 9 for n > k which I don't feel like doing right now, but which is relatively easy.
EDIT 2: What the heck, I'll do it anyhow.
We assumed that n > k. Therefore n - k > 0
Now n - k could be > k or it could be = k or < k. Let's try the first possibility first.
10^(n-k) - 10^k =
10^k (10^(n-2k) - 1)
Clearly, any power of 10 minus 1 (the second coefficient) is divisible by 1.
Now let's assume that n-k = k. In this case,
10^(n-k) - 10^k =
10^k - 10^k =
0
Clearly, 0 is divisible by 9.
Finally, let's assume that k > n-k.
10^(n-k) - 10^k =
10^(n-k)( 1 - 10^(2k-n))
As above, the second coefficient is divisible by 9.
QED