MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

I think, details are fairly simple at least in the case $a_n=n$. Indeed, if we must carry on is go up by $+n$, $+(n+1)$, $+(n+2)$, \dots, $+(n+k)$ we increase y-coordinate by $n+\dots+(n+k)$. We could replace $+s$ to $-s$, then we would get $2s$ less sum. So, by choosing $k$ and $s$ (we may use not only one $s$, but say 2 or three different values of $s$, but do not choose two consecutive $s$) we may get all numberssufficiently large $y$-coordinates, making at most 2 or 3 down jumps. AgainMoreover, choose even if $k$ is fixed we may get almost all coordinates of corresponding parity. Then we have the least not used number and do some work for placing it same freedom going to the right, then going down, and returning to the position left. Choose parities and jumps to "boundary opposite" directions for going to the necessary pointwith maximal used number".Well, this is not a rigorous proof, some technical details are really missed, but I do
For $a_n=n^2$ it should not expect difficulties herebe much harder.