1) Number the rows $1,2,...,2008$. Either all the odd rows or all the even rows will contain at least half of the square marked $0$ - w.l.o.g. suppose even rows. So in every odd row apply the operation to the squares in these rows until they become one region containing one number $(\le 1)$. 2) Every vertex in an even row is adjacent to an odd row. So take every square marked $1$ (in an even row) and apply the operation to it and the odd row above or below. 3) So far at least half of the $0$-marked squares haven't been operated on. Every square containing a $1$ has been operated on and are now a part of larger regions. Now we connect all of these larger regions. Take column (introducing at most $1004$ squares marked $0$) and apply the operation to all squares in the collumn. 4) We are left with one large region which contains some number $\le 1$ (because we've only been averaging numbers $\le 1$). There are also at least $\frac{1}{2}2008^2 - (\frac{1}{4}2008^2 +1004)>10^6,$ $\, 1\times 1$ squares marked $0$. Now in each remaining step we apply the operation to the larger region and a remaining $1\times 1$ square. Each step the value of the larger sqaure decreases by $\frac{1}{2}$. In the end, the number in the $2008\times 2008$ square is $< \frac{1}{2^{10^6}}\qquad\square$

Can you post the intution behind step 1?? I tried each of the steps 2,3,4 (i.e., trying to take care of all 1's saving at least half of the zeroes etc.) but to no avail. I understand that (1) is the crucial step. Nice problem and very nice solution by ocha!