Also 2014 Tournament of Towns Junior.

By the way, the bound can be improven to 1853, as one of the official solutions suggests.

Let me post the solution. I hope it's correct. We define an $m\times n$ grid have $m$ columns and $n$ rows. We cut the $100\times 100$ grid into $10$ pieces, with each piece being $10\times 100$. Then we cut each piece into $100$ slices, each being $10\times 1$. We consider the worst case, call it the "F is for Fear" scenario. In the F is for Fear, we expect to go through all $1000$ slices. If we move from slice to slice, it will take $1$ step if the two slices is in the same piece, or $10$ step if not. We want to minimize the steps here, so we only move $9$ times with the $10$ step case (it's always possible, even in the F is for Fear). So we will move $990$ times with the $1$ step case. Now in each slice, we would have to eat each food in the slice. In the F is for fear, we would need an average of $9$ step for eating each food. Summing up gives you $900$ steps. This would give you a total of $90+990+900=1980$ steps. PS: I think it can be generalized into all $n\times n$ if $n$ is a perfect square. It guarentees a total of no more than $(n-\sqrt{n})+(n-1)\sqrt{n}+n(\sqrt{n}-1)=(2n-2)\sqrt{n}$. But I don't know how to do it if $n$ is not a perfect square.