I changed people's names, sorry.

@OP just wanted to point out the title is Izho 2015, might want to edit that

InterLoop wrote: @OP just wanted to point out the title is Izho 2015, might want to edit that My bad , thank you <3

Here's what I managed to reduce the problem to: For $i, j\in\{1, 2, ..., 1001\}$, $i\neq j$, consider $m(i,j)$ the midpoint of the arc $(i,j)$ which is also a vertex of the $1001$-gon. Notice that $m(i,j)= \begin{cases} \dfrac{i+j}{2}\text{, when }i\equiv j\mod 2\\ \dfrac{i+j+1001}{2}\text{ (modulo }1001\text{), when }i-j\text{ is odd} \end{cases}$ So, in the end if there are $n$ vertices still white, then these form a set of points $A$ with the property that for each $i, j\in A$, $m(i,j)$ is also in $A$, and we would like to find the biggest $n$ for which such set exists. I noticed that any regulated $n$-gon works, with $n$ odd. That implies that $n\mid 1001$, so the biggest $n$ that works would be $11\cdot 13=143$, which means that the answer is $1001-143=858$. Unfortunately, I can't find a way to prove that $A$ is always the vertices of a regulated $n$-gon, but if someone manages to prove this, the solution is complete.

nooooo the title

Double07 wrote: Here's what I managed to reduce the problem to: For $i, j\in\{1, 2, ..., 1001\}$, $i\neq j$, consider $m(i,j)$ the midpoint of the arc $(i,j)$ which is also a vertex of the $1001$-gon. Notice that $m(i,j)= \begin{cases} \dfrac{i+j}{2}\text{, when }i\equiv j\mod 2\\ \dfrac{i+j+1001}{2}\text{ (modulo }1001\text{), when }i-j\text{ is odd} \end{cases}$ So, in the end if there are $n$ vertices still white, then these form a set of points $A$ with the property that for each $i, j\in A$, $m(i,j)$ is also in $A$, and we would like to find the biggest $n$ for which such set exists. I noticed that any regulated $n$-gon works, with $n$ odd. That implies that $n\mid 1001$, so the biggest $n$ that works would be $11\cdot 13=143$, which means that the answer is $1001-143=858$. Unfortunately, I can't find a way to prove that $A$ is always the vertices of a regulated $n$-gon, but if someone manages to prove this, the solution is complete. For the example you can just assume that if there are less than $858$ say $k$ then A can find one more white cell that he can turn to black and that is because $k\nmid1001$

This solution was found by me after the contest Clearly answer is $858=1001-1001/7$ Victor's strategy: let label the vertices from $1, 2 ... 1001$ then if $7\nmid t$, then one of the number $(t-k,t+k)$ dont divisible by $7$ for any $k$. Then if Victor chooses that number, no matter what Mary chooses, vertices that are divisible by 7 cannot be colored red. Mary's steategy: Let $t$ is any white vertices (let after finitely steps Mary can't marked the vertices red anymore). Then if the chip is at the vertice $a$, then Mary chooses $t-a$ if $a<t$ (otherwise chooses $a-t$). Then Victor must choose counter clockwise direction (clockwise if $a>t$), because unless, the chip will be at vertice $t$, so the vertice $t$ become red, so Mary can place the chip on vertices $2a-t$ regardless of Victory's move. Let $t$ and $k$ are white vertices, then we can choose $t=1001$, because we labeled the vertice randomly. Then Mary can place the chip on vertices $2a$ or $2a-k$ from vertice $a$. So after $n$ step Mary can place the chip on vertice $2^na-(2^{x_1}+2^{x_2}+...+2^{x_j})k$ for all $n$ and $0\leq {x_1}<{x_2}<...<{x_j}\leq n$. So if $(k,1001)=1$ then for any integer $A$ we can find $0\leq {x_1}<{x_2}<...<{x_j}\leq n$ such that $2^na-(2^{x_1}+2^{x_2}+...+2^{x_j})k\equiv A \mod 1001$ (because we can write $\frac{2^na-A}{k}$ some degree of $2$'s). So Mary can color any vertice, if $(k, 1001)=1$ If $(k, 1001)>1$ then $(k-t, 1001)=(k-1001, 1001)=(k, 1001)\geq7$, so for any white vertices $k$ and $t$, their difference is at least $7$, so maximum value of white vertices is $\frac{1001}{7}=143$, then the minimum value of red vertices is $1001-143=858$.

The answer is $1001-143=\boxed{858}$. Instead of a regular $n=1001$-gon, we consider all $n$ residues in mod $n$. In each turn, Rose pick a number, then Brunno picks its sign. Assume that the initial position of the chip is $1$. Brunno's strategy. He firstly colors all multiples of $7$ residues blue. I claim that he can make sure no blue residue will ever be recolored red. His strategy is just to avoid landing on blue residue if possible. Suppose he cannot do so, then there is a moment that the chip is equally far from two distinct blue residues. Since both blue residues are $0\pmod 7$, the chip has to be $\equiv (0+0)/2\equiv 0\pmod 7$. This means the chip is already on a blue residue in the first place, which is a contradiction. Therefore, he can avoid coloring red on $1001/7=143$ residues, hence the best Rose can do is at most $858$. Rose's strategy. Suppose her best strategy can color at most $M$ residues red. Color all not-red residues in blue. Also, after we have colored all those $M$ residues red, suppose the counter is now at $s$. Call a residue $r$ good when there is a series of (Rose's) moves that ends at $r$ no matter how Brunno responses, otherwise, call it bad. I claim that there are at most $143$ bad residues. Claim. If $a$ and $b$ are bad, $\frac{a+b}{2}\pmod n$ is also bad. Proof. Suppose $\frac{a+b}{2}\pmod n$ is good, and Rose pick $k=\frac{a-b}{2}$. Then, either $a$ or $b$ has to be good, which is a contradiction. $\square$ Claim. For any distinct bad $a$, $b$, if $d=\gcd(a-b,n)$, then all $c\equiv a\pmod d$ are bad. Proof. Keeping in mind that $n$ is odd, we can use the previous claim iteratively all \[c\equiv \frac{ta+(2^k-t)b}{2^k}\equiv b+\frac{a-b}{2^k}t\pmod n\]are bad for any $k\in \mathbb{N}$ and odd positive $t<2^k$. Setting $k$ large enough and varying $t$ gives our desired result. $\square$ Let $u$, $v$ two closest bad residues and let $d=\gcd(u-v,n)$. Then, all $n/d$ residues of the form $w\equiv u\pmod d$ are bad, call these residues clean. Suppose there is another bad residue $r$ which is not clean. Then, $r$ is closer to a clean residue than $|u-v|$, which is absurd. Because at least one residue is good, $d\neq 1$. Therefore, as $7$ is the smallest factor $>1$ of $1001$, there are at most $1001/7=143$ bad (hence blue) residues. $\blacksquare$

Titibuuu wrote: I changed people's names, sorry. You got my hopes up there for a moment

v_Enhance wrote: Titibuuu wrote: I changed people's names, sorry. You got my hopes up there for a moment