if I got it correctly,the problem seems to be too simple. As it traditionally happens,we assume the contrary,so after some "jump" we get the described situation, and let's make all jumps in a reverse order, but as it is easy to observe that starting from that position,any distances between frogs must be divisible by $ 2008$,contradiction.

Another solution - use invariants. Colour the points on x (with integer coordinates) black, white, black, white... When we start 2 frogs are on white and 2 on black points. We see that after every jump the color of the frog point does not change, but if all distances are equal to 2008 all four points ale coloured the same - contradiction.

My solution much uglier than above two Assume contrary.Denote points in line as A,B,C,D in turn.Choose any point in our line as origin of coordinates.Denote coordinates of A,B,C,D after n-th "jumping" as $ a_n$,$ b_n$,$ c_n$,$ d_n$ respectively.denote $ l_n$ as sum of distances between A and B,B and C,C and D after n-th "jumping".$ l_0=3$.Notice that $ l_n\equiv b_n-a_n+c_n-b_n+d_n-c_n \equiv d_n-a_n\mod 2$.and it is easy to understand that $ d_n-a_n\equiv d_{n+1}-a_{n+1}\mod2$,so as we can see $ l_n \equiv l_0\mod2$.We get contradiction.