I claim that the answer is $\boxed{2}$ Firstly, let's show the bound. Color all columns in red and green. Let $W=(U,V)$ be an array of integers, where $U$ is the number of white pawns on red columns, and $V$ is the number of white pawns on green columns. Similarly, $B=(X,Y)$, where $X$ is for red columns and $Y$ is for green columns. We will view data in a table $\begin{tabular}{ c c } U & V \\ X & Y \end{tabular}$ Each move either nothing changes, or in some diagonal numbers decrease by $1$, and some other number increases by $1$. Suppose we can get $1$ pawn remaining, wlog the table would be $\begin{tabular}{ c c } 1 & 0 \\ 0 & 0 \end{tabular}$ We start with $\begin{tabular}{ c c } 50 & 50 \\ 50 & 50 \end{tabular}$ Consider $U+Y-X-V$. This number never changes mod $3$, however we start with $50+50-50-50=0$ and finish with $1+0-0-0=1$. Contradiction Enumerate pawns from left to right with numbers $1 \ldots 100$. Give cells coordinates, where white pawn $1$ is in $(1,1)$ Let black pawn $1$ eat white pawn $2$ at $(2,2)$, and then white pawn $1$ eats it. Similarly, white pawn $1$ will eat on $(3,3), \ldots,(99,99)$. Now just eat with eat the black pawn on $(100,100)$, and black pawn $99$ eats white pawn $100$

This problem killed me on contest... Also, I've heard from coordinators that no solution using induction or geometric arguments has been found so far. Only this invariant. Someone get on that?

Actually, at the end it seems two other approaches possibly can work (90% sure according to coordinators) *considering number of captures between column i and i+1 (if 1 pawn remains; 98 times 2 and 1 times 3) And doing case study for what happens when there are 3 captures *induction and many many cases (So not elegant at all) Essentially, there is only one elegant proof indeed (up to way of working out)

I would really like to see a full writeup of such a solution as I've spent days trying to complete those ideas with several other students at the camp to no avail.

Well, I'm not sure what you call "other solution", but initially my proof in the post finished like that: let $x_1,x_2,x_3,x_4$ be the number of types of each of the changes of the table ($x_1$ - when top-left increases, $x_2$ - top right, etc). Then we have a system of four equations on this variables, and when you solve it, it turns out that the solution is not integral, giving the desired contradiction

gvole wrote: I would really like to see a full writeup of such a solution as I've spent days trying to complete those ideas with several other students at the camp to no avail. Sketch of solution using the number of captures between columns $i$ and $i+1$. $Step$ $1.$ The number of pawns on each column cannot go up $Step$ $2.$ Looking at columns $i$ and $i+1$ you have at most $3$ captures $Step$ $3.$ The number of captures is $2n - k$, where $k$ is the number of pawns left at the end $Step$ $4.$ If you have $3$ captures between $i$ and $i+1$, then somewhere must be a $1$ $Step$ $5.$ Assume you can have only one pawn at the end. Prove that then there cannot be $1$s $Step$ $6.$ Because we have $2n-1$ captures, we must have some index $i$ such that between columns $i$ and $i+1$ we have $3$ captures, so we must have a $1$ somewhere, contradiction So, we have at least $2$ pawns at the end.

This problem is wrongly credited: it was proposed by me, José Alejandro Reyes González (from Mexico as well). I am not smart enough to understand if the solution above is correct, nor am I good enough at combinatorics to know if a solution like the one described is possible. An interesting variant of the problem is as follows (which I have not been able to fully solve): Ana plays a game on a $\color{red}100\times k$ chessboard. Initially, there is a white pawn on each square of the bottom row and a black pawn on each square of the top row, and no other pawns anywhere else, for a total of 200 pawns. Each white pawn moves toward the top row and each black pawn moves toward the bottom row in one of the following ways: it moves to the square directly in front of it if there is no other pawn on it; it captures a pawn on one of the diagonally adjacent squares in the row immediately in front of it if there is a pawn of the opposite color on it. (We say a pawn $P$ captures a pawn $Q$ of the opposite color if we remove $Q$ from the board and move $P$ to the square that $Q$ was previously on.) Ana can move any pawn (not necessarily alternating between black and white) according to those rules. What is the smallest number of pawns that can remain on the board after no more moves can be made?