Surely odd numbers cannot remain as the final number as the sum is invariant mod 2. 

Now, we show that every even number is attainable. For $1 \leq k \leq 2^7$,

$$(1,2k), (2,3), (4,5), \ldots, (2k-2,2k-1), (2k+1,2k+2), \ldots, (2^{8}-1, 2^{8}) $$

$$\Rightarrow 2k-1, 1, 1, \ldots, 1 \qquad \text{(note that there are an odd number of 1's)} $$

$$\Rightarrow (2k-1, 1), (1,1), (1,1), \ldots, (1,1) $$

$$\Rightarrow 2k-2, 0, 0, \ldots, 0 $$

$$\vdots $$

$$\Rightarrow 2k-2.$$