Represent such an arrangement as a $100\times 100$ matrix $B$. Where entry $b_{i,j}=1$ iff there is a king in the $i$th row and $j$th column of the chessboard. In addition, let $r_i=(b_{i,1},\hdots, b_{i,100})$ and $c_j=(b_{1,j},\hdots, b_{100,j})$. Lemma: If $i$ is odd, $r_i+r_{i+1}$ equals either $(0,1,0,\hdots, 1)$ or $(1,0,1,\hdots, 0)$. If $i$ is even, $r_i+r_{i+1}$ equals either $(0,1,0,\hdots, 1)$, $(1,0,1,\hdots, 0)$, or $(1,0,\hdots,1,0,0,1,0,\hdots,1)$ where there are $25$ $1$'s on each side of the pair of $0$'s. Proof: If $i$ is odd, since no two kings in the same row can be in the same column, no entry of $r_i+r_{i+1}$ is $2$. Also, no two $1$'s of $r_i+r_{i+1}$ can be adjacent. Thus, if $r_i+r_{i+1}$ is not one of those two vectors, then there exist two consecutive zeros in $r_i+r_{i+1}$. Thus, there exists a $j$ such that \[\begin{bmatrix}b_{ij}&b_{i,j+1}\\b_{i+1,j}&b_{i+1,j+1}\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\] Therefore, the kings in columns $j$ and $j+1$ are in $(i-1)\times 2$ and $(99-i)\times 2$ rectangles. Since $i$ is odd, the two rectangles can holds at most \[\left\lfloor \frac{i-1+1}{2}\right\rfloor+\left\lfloor \frac{99-i+1}{2}\right\rfloor=49\] kings. This contradicts that the two columns hold $50$ kings. Hence, $r_i+r_{i+1}$ equals either $(0,1,0,\hdots, 1)$ or $(1,0,1,\hdots, 0)$ if $i$ is odd. If $i$ is even, then we know that $r_i+r_{i+1}$ equals either $(0,1,0,\hdots, 1)$, $(1,0,1,\hdots, 0)$, or $(1,0,\hdots,1,0,0,1,0,\hdots,1)$ with an arbitrary number of $1$'s on each side of the $0$'s. From the $i$ is odd case, we know that the parity of the $j$ for which $b_{i,j}=1$ is constant for fixed $i$. Therefore the number of $1$'s on each side of the pair of $0$'s in $(1,0,\hdots,1,0,0,1,0,\hdots,1)$ is $25$, as desired. $\boxed{}$. Since there must exist a king in every column, there must exist two consectutive rows $r_i$ and $r_{i+1}$ which sum to $(1,0,\hdots,1,0,0,1,0,\hdots,1)$. Thus, every row must be one of the following vectors with $25$ ones. \[v_1=(1,0,\hdots, 1,0,0,0,\hdots, 0,0)\\ v_2=(0,1,\hdots, 0,1,0,0,\hdots, 0,0) \\v_3=(0,0,\hdots, 0,0,1,0,\hdots, 1,0)\\ v_4=(0,0,\hdots, 0,0,0,1,\hdots, 0,1)\] By symmetry, $c_j$ must also be one of those vectors. If $r_1=v_1$, then $r_2=v_3$ and $r_3=v_1$. So $r_i=v_1$ for all odd $i$, contradiction. We get a similar contradiction if $r_1=v_4$. If $r_1=v_2$, then $r_2=v_4$. Thus, $c_j=v_1$ if $j\le 50$ is an even number and $c_j=v_2$ if $j>50$ is an even number. By the analogous lemma for columns, $c_j=v_3$ if $j\le 50$ is an odd number and $c_j=v_4$ if $j>50$ is an odd number. The chessboard corresponding to $B$ can be verified to have the desired properties. If $r_1=v_3$, we get a similar arrangement. Thus, there are exactly $2$ arrangements.

http://wiki.logic-masters.de/index.php?title=Starbattle/en Not the most original problem. Although I suppose there are not too many people in the intersection of math olympiads and competitive logic puzzling (me included since I made a transition in between), it's a well-known tidbit of information for most World Puzzle Championship competitors that 8 by 8 Star Battle puzzles have only two possible solutions even before looking at the regions. On learning this a long time ago I went ahead and proved it and the generalization. So I had done this problem long before ever seeing the IMO shortlist.

Easy solution: Call a king a "flower king" if it does not attack anybody. Divide the board into 2500 2x2 boards. It is trivial that each one has exactly one king. Now, take any "double column", a column composed of 50 2x2 boards. Call one of these boards UP if its king is one of the two upper squares, and DOWN if not. Suppose not all boards are of the same type. Since (counting from the upper to the lower board) we cannot have a DOWN board followed by an UP board, we must have an UP board followed by a DOWN board. This leaves two rows in between with no kings in the double column. We have 98 2x2 boards containing these rows, 49 upper ones and 49 lower ones (ignoring the ones in the double column). From the 49 upper ones, at least 25 are DOWN boards and from the lower ones, at least 25 are UP boards so we have two consecutive boards, the upper one is DOWN and the lower one is UP. Contradiction! Therefore, all the 2x2 boards from each double column are of the same type. We call this the type of the double column, so there can be UP double columns and DOWN double columns. The same applies for RIGHT double rows and LEFT double rows. Suppose we have some consecutive UP double columns, which have LEFT double columns surrounding them. I will prove a contradiction. The argument I will make works anyway, so assume we have TWO UP double columns surrounded by LEFT double columns. Take the two (single) columns in between. They have 25 kings each. Therefore, from the 50 leftmost 2x2 boards, 25 are LEFT and 25 are right, similarly from the 50 rightmost boards, 25 are LEFT and 25 are RIGHT. Now, WLOG take a LEFT board $B$ from the left, which has a RIGHT board on top (either this happens or there is a RIGHT board from the right with a LEFT board on top. They are the same case). Now take the double column to the left (this double columns is a DOWN double column). We can see that the boards adjacent to the respective LEFT boards must be LEFT boards too, so there are 25 LEFT boards (adjacent to the other LEFT boards) and 25 RIGHT boards (adjacent to RIGHT boards). We can easily see the king in $B$ is attacking the king on its upper-right corner, contradiction. Therefore, either the double columns alternate UP-DOWN-UP-DOWN...-UP-DOWN, or there are only 2 blocks: UP-UP-...-UP-DOWN-...-DOWN (and because every row has 25 kings, there would be 25 UP double columns and then 25 DOWN double columns). (The same occurs for the double rows). Assume the first case, that they alternate. Assume the upper-left board is a LEFT board. Therefore that double row is a LEFT double row, and the 2º board is also a LEFT board, so that its king is located in the lower-left corner. Since it cannot attack the king on the board below the upper-left board, the upper let board must be LEFT too. Therefore, the double rows do not alternate, they are LEFT-LEFT-...-LEFT-RIGHT-...-RIGHT. So we know the configuration of the board, and we can easily reach a contradiction. Now, if the upper-left board is a RIGHT board, the first doublerow is a RIGHT doublerow. By looking at the kings on the boards adjacent to the board that is to the left of the upper-right board, we can see the second double row is RIGHT also, so the doublerows do not alternate. Therefore, we can fill the whole board and easily reach a contradiction, looking at the center of the board. Therefore the doublecolumns are UP-...-UP-DOWN-...-DOWN or DOWN-...-DOWN-UP-...-UP, and the same for the doublerows. We can easily see that only 2 of these configurations work: - UP-...-UP-DOWN-...-DOWN and RIGHT-...-RIGHT-LEFT-...-LEFT - DOWN-...-DOWN-UP-...-UP and LEFT-...-LEFT-RIGHT-...-RIGHT These configurations, lets call them the "Flower Kings" configurations, work, so the answer is $\boxed{2}$.

Divide the board into $2\times 2$ square sections (so a $50\times 50$ grid of $2\times 2$ squares). Note that by Pigeonhole, if one of these $2\times 2$ squares is blank, then there exists another $2\times 2$ square with $2$ kings in it, contradiction. Thus, every $2\times 2$ square must have exactly one king. Now label each $2\times 2$ square with either a $L$ or $R$, and either a $U$ or $D$, to denote which of the four squares the king is on. Note that for every row of fifty $2\times 2$ squares, exactly $25$ are $U$ and $25$ are $D$. In addition, note that it is impossible for a square below a $D$ square to be $U$, or else clearly the two kings will attack. Thus, it follows that all fifty $2\times 2$ squares of any column are either all $D$ or all $U$, else if there exists a switch from $U$ to $D$ there also exists a switch from $D$ to $U$ on the same pair of adjacent rows, which we already know cannot happen. By a similar argument, all fifty $2\times 2$ squares of any row are either $R$ or $L$. Thus, instead of labelling each $2\times 2$ square, we can just label each $2\times 2$ row and column with $L/R$ and $D/U$'s respectively. In addition, we label exactly $25$ rows/columns with $L, R/D, U$ because of the last condition. Now let's consider a $4\times 4$ square formed by two adjacent rows and two adjacent columns. If the columns are $D, U$ from left to right and the rows are $R, L$ from top to bottom, then the top left and bottom right $2\times 2$ squares' kings will attack each other. Similarly, if the columns are $U, D$ and the rows are $L, R$, the top right and bottom left kings will attack each other. Thus, we can only have both the transition $D\to U$ and $L\to R$ or $U\to D$ and $R\to L$. Since these transitions must happen in the middle two adjacent rows and columns because exactly $25$ rows/columns are labeled with $L, R/D, U$, there are no more ways to permutate them and so these are the only two ways to place the kings. The answer is $\boxed{2}$. EDIT: sorry, I did not realize my solution is the same as antimonyarsenide's solution.

An interesting variant is to get rid of the second condition. The number of possibilities is made quite higher than 2

GoldenFrog1618 wrote: Represent such an arrangement as a $100\times 100$ matrix $B$. Where entry $b_{i,j}=1$ iff there is a king in the $i$th row and $j$th column of the chessboard. In addition, let $r_i=(b_{i,1},\hdots, b_{i,100})$ and $c_j=(b_{1,j},\hdots, b_{100,j})$. Lemma: If $i$ is odd, $r_i+r_{i+1}$ equals either $(0,1,0,\hdots, 1)$ or $(1,0,1,\hdots, 0)$. If $i$ is even, $r_i+r_{i+1}$ equals either $(0,1,0,\hdots, 1)$, $(1,0,1,\hdots, 0)$, or $(1,0,\hdots,1,0,0,1,0,\hdots,1)$ where there are $25$ $1$'s on each side of the pair of $0$'s. Proof: If $i$ is odd, since no two kings in the same row can be in the same column, no entry of $r_i+r_{i+1}$ is $2$. Also, no two $1$'s of $r_i+r_{i+1}$ can be adjacent. Thus, if $r_i+r_{i+1}$ is not one of those two vectors, then there exist two consecutive zeros in $r_i+r_{i+1}$. Thus, there exists a $j$ such that \[\begin{bmatrix}b_{ij}&b_{i,j+1}\\b_{i+1,j}&b_{i+1,j+1}\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}\]Therefore, the kings in columns $j$ and $j+1$ are in $(i-1)\times 2$ and $(99-i)\times 2$ rectangles. Since $i$ is odd, the two rectangles can holds at most \[\left\lfloor \frac{i-1+1}{2}\right\rfloor+\left\lfloor \frac{99-i+1}{2}\right\rfloor=49\]kings. This contradicts that the two columns hold $50$ kings. Hence, $r_i+r_{i+1}$ equals either $(0,1,0,\hdots, 1)$ or $(1,0,1,\hdots, 0)$ if $i$ is odd. If $i$ is even, then we know that $r_i+r_{i+1}$ equals either $(0,1,0,\hdots, 1)$, $(1,0,1,\hdots, 0)$, or $(1,0,\hdots,1,0,0,1,0,\hdots,1)$ with an arbitrary number of $1$'s on each side of the $0$'s. From the $i$ is odd case, we know that the parity of the $j$ for which $b_{i,j}=1$ is constant for fixed $i$. Therefore the number of $1$'s on each side of the pair of $0$'s in $(1,0,\hdots,1,0,0,1,0,\hdots,1)$ is $25$, as desired. $\boxed{}$. Since there must exist a king in every column, there must exist two consectutive rows $r_i$ and $r_{i+1}$ which sum to $(1,0,\hdots,1,0,0,1,0,\hdots,1)$. Thus, every row must be one of the following vectors with $25$ ones. \[v_1=(1,0,\hdots, 1,0,0,0,\hdots, 0,0)\\ v_2=(0,1,\hdots, 0,1,0,0,\hdots, 0,0) \\v_3=(0,0,\hdots, 0,0,1,0,\hdots, 1,0)\\ v_4=(0,0,\hdots, 0,0,0,1,\hdots, 0,1)\] By symmetry, $c_j$ must also be one of those vectors. If $r_1=v_1$, then $r_2=v_3$ and $r_3=v_1$. So $r_i=v_1$ for all odd $i$, contradiction. We get a similar contradiction if $r_1=v_4$. If $r_1=v_2$, then $r_2=v_4$. Thus, $c_j=v_1$ if $j\le 50$ is an even number and $c_j=v_2$ if $j>50$ is an even number. By the analogous lemma for columns, $c_j=v_3$ if $j\le 50$ is an odd number and $c_j=v_4$ if $j>50$ is an odd number. The chessboard corresponding to $B$ can be verified to have the desired properties. If $r_1=v_3$, we get a similar arrangement. Thus, there are exactly $2$ arrangements. Hello I don't think the second part of the lemma is true.Can somebody justify it.

Basically the same solution as some of the ones above, but I'm still gonna post because I've got some Pretty Pictures${\texttrademark}$, now with 25% extra pictures! Albeit it's not necessary for the proof, it's intersting to note that the bound of $2500$ is tight. To see why, divide the board into $2\times 2$ subgrids, and note that each of the $2500$ subgrids can have at most one king. While not strictly necessary for the problem at hand, it gives the motivation to: Divide the grid into $2\times 2$ subgrids and note that each subgrid can have at most $1$ king. Since there are $2500$ subgrids, each of them must have exactly one king. Now in each sub-grid, there are four possible positions for the king. We'll label the correspoding subgrids $\text{LU, RU, LD, RD}$ as shown: [asy][asy]size(6cm); usepackage("skak"); for(int i=0;i<3;++i){draw((0,i)--(2,i));draw((i,0)--(i,2));} for(int i=0;i<3;++i){draw((3,i)--(5,i));draw((i+3,0)--(i+3,2));} for(int i=0;i<3;++i){draw((6,i)--(8,i));draw((i+6,0)--(i+6,2));} for(int i=0;i<3;++i){draw((9,i)--(11,i));draw((i+9,0)--(i+9,2));} label("$\symking$",(0.5,1.5));label("$\symking$",(4.5,1.5));label("$\symking$",(6.5,0.5));label("$\symking$",(10.5,0.5)); label("LU",(1,0),S);label("RU",(4,0),S);label("LD",(7,0),S);label("RD",(10,0),S); [/asy][/asy] Consider the leftmost column made of $2\times 2$ subgrids (so, a $100\times 2$ thingy). The second column has exactly $25$ kings, so $25$ of those $50$ subgrids is of $\text{R\#}$ type (here $\#$ can be $\text{U}$ or $\text D$). Now we observe that for any sub-grid of $\text{R\#}$ type, the subgrid to its right has to be of $\text{R\#}$ type as well. So all the subgrids that are in the same row as one of those $25$ subgrids are of the same type, giving us $25$ all-$\text R$ rows. Similarly, the other $25$ rows are $\text L$ rows. In an analogous fashion, we can prove $25$ of the subgrid columns are $\text U$ and others are $\text D$. [asy][asy]size(7cm);usepackage("contour", "outline"); texpreamble("\contourlength{1.7pt}"); draw((0,0)--(0,-5)--(5,-5)--(5,0)--cycle); draw((0,-1)--(5,-1)^^(0,-2)--(5,-2)^^(1,0)--(1,-5)^^(2,0)--(2,-5)); draw((0,-0.5)--(5,-0.5)^^(0,-1.5)--(5,-1.5)^^(0.5,0)--(0.5,-5)^^(1.5,0)--(1.5,-5),dotted); label("R"+"$\rightarrow$",(0,-1.5),W);label("L"+"$\rightarrow$",(0,-0.5),W); label("\contour{white}{RD}",(0.5,-1.5));label("\contour{white}{RU}",(1.5,-1.5)); draw(brace((-1,-5),(-1,0)));label(rotate(90)*"25 L's,25 R's",(-2,-2.5)); draw(brace((0,0.2),(5,0.2)));label("25 U's, 25 D's",(2.5,1.2)); [/asy][/asy] Now suppose (i) there is some $\text L$ subgrid row directly below some $\text R$ subgrid row, and (ii) there is some $\text U$ subgrid column directly to the right of some $\text D$ subgrid column. Then the following situation arises: [asy][asy]usepackage("skak");size(6cm);draw((0,0)--(0,-5)--(5,-5)--(5,0)--cycle); draw((0,-2)--(5,-2)^^(2,0)--(2,-5)); label("R",(0,-1.8),W);label("L",(0,-2.2),W); label("D",(1.8,0),N);label("U",(2.2,0),N); label("$\symking$",(2,-2),NW);label("$\symking$",(2,-2),SE); [/asy][/asy] This has two attacking kings, a bad thing. So at least one of (i) and (ii) doesn't hold; say (i). That means the subgrid rows are $\underbrace{\text{LLL...L}}_{25}\underbrace{\text{RRR...R}}_{25}$ in that order. In this case, if there's a $\text D$ column directly after one $\text U$ column, the situation becomes this:[asy][asy]usepackage("skak");size(6cm);draw((0,0)--(0,-5)--(5,-5)--(5,0)--cycle); draw((0,-2)--(5,-2)^^(2,0)--(2,-5)); label("L",(0,-1.8),W);label("R",(0,-2.2),W); label("U",(1.8,0),N);label("D",(2.2,0),N); label("$\symking$",(2,-2),NE);label("$\symking$",(2,-2),SW);[/asy][/asy]which is again bad. So the columns are $\underbrace{\text{DDD...D}}_{25}\underbrace{\text{UUU...U}}_{25}$. This gives a unique configuration of kings. Similarly, the case where (ii) doesn't hold gives another unique configuration, so there are $\boxed{2}$ of them in all. Just for fun, here is one of the configuartions for a $8\times 8$ board (the other one is simply a reflection of this): [asy][asy]size(5cm);usepackage("skak"); for(int i=0;i<9;++i){draw((i,0)--(i,8)^^(0,i)--(8,i));} pair[] a={(2,1),(4,1),(2,3),(4,3),(1,5),(3,5),(1,7),(3,7),(6,2),(8,2),(6,4),(8,4),(5,6),(7,6),(5,8),(7,8)}; for(pair k:a){label("$\symking$",k+(-0.5,-0.5));} [/asy][/asy]

Split the graph into a 50x50 lattice of 2x2's. Since there are 2500 kings, and at most 1 king per 2x2, easy to see that there are exactly 1 king in each 2x2. We see that if: K is king ______ |__|__| |__|_K| Then all 2x2's to the right of it are(same row): ______ |__|__| |__|_K| or ______ |__|_K| |__|__| Thus each row and column has a divider that divides the 2x2's that look like 2 above and the ones that are not the 2 above. It is easy to see that any 2x2 is uniquely defined by it's divider. Also for every row, there are exactly 25 column dividers above and 25 below.(Since 25 kings in each row) Thus we see that the dividers must be on the sides of the board, 25 on each side. But, we cannot have: _____ |__|__| |__|__| Note: In this graph every line has length 50, reds are dividers. Thus we see that there are only 2 ways to arrange the dividers.(We can reflect it) Sry for bad diagrams

We use 1 to represent occupied squares and 0 otherwise. Let $r_i$ be the $i$th row (which is a 100-element list, all zeros and ones), and let $s_i=r_i\vee r_{i+1}$. Then each $s_i$ must have exactly 50 ones and at least one zero between any two adjacent ones. There is a set $L$ of exactly 25 rows that have a "1" on its intersection with the second column, so $s_i$ and $s_{i-1}$ are both $(0,1,0,1,\dots,0,1)$ for any $r_i \in L$. Similarly there is a set $R$ of exactly 25 rows that have a "1" on its intersection with the second-to-last column, so $s_i$ and $s_{i-1}$ are both $(1,0,1,0,\dots,1,0)$ for any $r_i \in R$. Lemma: It can't happen that $s_i=(0,1,\dots,0,1)$ and $s_{i+1}=(1,0,\dots,1,0)$. Proof: Simply observe that $s_i \wedge s_{i+1}$ is all zero, so $r_{i+1}$ is all zero, which is impossible. Because $|L|=25$, there are at least 49 $s_i = (0,1,\dots,0,1)$ and at least 49 $s_j = (1,0,\dots,1,0)$. (Here one needs to verify that the first and last rows can't both be in $L$: otherwise there will be at least 48 $s_i = (0,1,\dots,0,1)$ and 50 $s_j = (1,0,\dots,1,0)$, and there will be at least two more $s_k$ not belonging to either of them by the lemma, which is impossible.) So we conclude that there are exactly 49 of each that respectively form consecutive blocks, plus one $s_k$ on the boundary of the two blocks. Without loss of generality, suppose $s_1$ through $s_{49}$ are $(0,1,0,1,\dots,0,1)$, and $s_{51}$ through $s_{99}$ are $(1,0,1,0,\dots,1,0)$. To finish, we will focus on this $s_{50}$. Because $s_{50}\wedge s_{49}$ and $s_{50}\wedge s_{51}$ both need to have at least 25 ones, the only possible $s_{50}$ is $(1,0,\dots,1,0,0,1,\dots,0,1)$ with the two consecutive zeros in the exact middle. This determines the entire state of the 100-by-100 grid. Because we have WLOGed one case out of two, we conclude that there are two valid arrangements in total.

Split the chessboard into $2500$ $2\times 2$ cells, so that no two cells overlap. Then, if some $2 \times 2$ cell contains no king, by pigeonhole some other $2 \times 2$ cell contains $2$ kings, a contradiction as these would attack each other. Hence, each $2 \times 2$ cell contains exactly $1$ king. To each $2 \times 2$ cell, we assign an ordered pair $(X, Y)$, where $X \in \{L, R\}$ and $Y \in \{D, U\}$. An assignment of the pair $(L, D)$ signifies that the king in some given cell is in the bottom left square. Define the \textit{signature} of a row of cells to be the string consisting of the $y$-coordinates of the cells in the row in order, and define the signature of a column of cells similarly. The condition of $25$ kings per row/column implies that each row signature contains $25$ $U$'s and $25$ $D$'s, while each column signature contains $25$ $R$'s and $25$ $L$'s. Main Claim: Every row has the same signature, and every column has the same signature. Proof: We prove only that all rows have the same signatures; the proof for columns is identical. Suppose the bottom row has signature $r_1r_2\cdots r_{50}$, and suppose the second row has signature $s_1s_2\cdots s_{50}$. Note that if $r_i$ is a $U$, $s_i$ must also be a $U$ (if $s_i$ was a $D$, there would be two kings attacking each other). Hence, as $25$ of the $r_i$'s are $U$'s, the $25$ corresponding $s_i$'s are also $U$. However, exactly $25$ of the $s_i$'s are $U$'s, implying that if $r_i$ is a $D$, we must also have $s_i$ being a $D$. Therefore, the bottommost row has the same signature as the second row; repeating this argument implies that all rows have the same signatures. $\blacksquare$ Now, suppose the row signature (as there is only one) contains both $UD$ and $DU$ as substrings. The column signature contains at least one $LR$ and $RL$ as a substring. If there is an $RL$ substring, the $UD$ and $RL$ substringstogether give two diagonally touching cells (where the diagonal is parallel to $y = x$) which are assigned pairs $(R, U)$ and $(L, D)$; the kings in these cells attack each other, a contradiction. A similar contradiction is reached with an $LR$ substring, meaning that the row signature cannot contain both $UD$ and $DU$ as substrings. Similarly, the column signature cannot contain both $RL$ and $LR$ as substrings. Thus, the row signature is one of $U\cdots UD\cdots D$ and its reverse, while the column signature is one of $R\cdots RL\cdots L$ and its reverse. If the row signature is $U\cdots UD \cdots D$, the column signature cannot contain a $RL$ substring, implying that the column signature is $L \cdots L R\cdots R$; similarly, if the row signature is $D \cdots DU \cdots U$, the column signature is $R \cdots RL\cdots L$. These assignments of signatures obviously give working arrangements, implying that the answer is $\boxed{2}$. $\Box$

I hope my “solution” is correct. Anyway I loved this. First off, to give some structure to the problem, we divide the board into $2 \times 2$ blocks and note that each of them must have exactly $1$ king in it. We classify the blocks into UR (up right), UL(up left), DR(down right), DL(Down left) depending on the position of the king. Also when we say a block is $R$ type and so on, it is $RU$ or $RD$. Also we let rows be defined as $2 \times 100$ pieces. Claim 1 : Any row has either $U$ type or $D$ type blocks. Proof : suppose not. We note that if the top most and bottom most block are $U$ or $D$, the whole column is $U$ and $D$ respectively. So FTSOC it’s possible that the topmost block and bottom most are a different type. The only possibility is that the topmost block is $U$ and bottom most block is $D$ for obvious reasons. Let there be $a$ instances of this. Let there be $b$ instances of them being the same type too. We count the number of kings in the second row (row defined in the normal sense here) as $b$, and the number of kings in the last row (defined in the normal sense again) to be $a+b$. Clearly we have $a+b=b \implies a=0$. So we are done $\blacksquare$. Claim 2 : ALL the blocks in the same row are either $R$ or $L$ type. Proof: $R$ types always produce $R$ types $\blacksquare$ Claim 3 : $DR$ can never be above $DL$. Proof : By claim 1, if the king in one box is shifted upwards, all the kings in the row are. Take the first instance of this translation (which must exist for obvious reasons) and we see two kings in the neighbourhood of each other. Very similarly we can prove that $UL$ always has to be above $UR$. Claim 4: DR s and DL s are together in a column. The same holds for ULs, URs Proof : we see that at some point $DR, DL$ must be neighbours. By claim 1 we know that the whole grid is filled with rows being the copies of the first one, or the first row with kings translated upwards. At some point the row must be translated ; we take a look at the adjacent $U,D$ types. By assuming the contrary one can easily see that two kings are neighbours in that case $\blacksquare$. Claim 5 : all the U type and D type rows are together. Proof : consider the neighbour rows which are of different types. Again, suppose the contrary (that there exists something of this type : UDU or DUD). Making the diagram and considering the points where the $L, R$ boxes are neighbours, again we see that two kings are neighbours, so we get a contradiction This all implies we have $\boxed{2}$ configurations possible. P.S. I apologise If I messed up rows and columns somewhere, and for no asymptote

The answer is $2.$ Divide the chessboard into 2500 $2\times 2$ subgrids. Note that each box can have at most one king, implying each box must in fact have exactly one king. Refer to the king in the box of the $i$th row and $j$th column as $B_{i,j}$ where $1\le i,j\le 50.$ Refer to a layer as a row of $n$ of these $2\times 2$ boxes and a wall as a column of these $2\times 2$ boxes. For each box $b_1, b_2,\dots, b_{50}$ of the leftmost wall, we can label it as either $L$ or $R$ to indicate which of the two cell columns each king of the given box $b_i$ lies on (left or right). Similarly, for each box of the topmost layer, we can label each box as either $T$ or $B$ to indicate whether the king lies in the top or bottom cell row of the layer. Consequently, for a given box, we can uniquely determine which of the 4 cell positions the king lies as either $(L,T), (L,B), (R,T), (R,B).$ Claim: If the topmost layer has labelling $a_1, a_2,\dots, a_{50}$ where $a_i\in \{T,B\},$ then every subsequent layer must follow the same labelling. Similarly for leftmost wall and subsequent walls. Proof. Note that for the topmost layer, 25 of them must have $T$ and 25 must have $B.$ Let $a_{r,i}$ denote the label of the $i$th box of the $r$ layer. Then, if $a_{r,i} = B$ then $a_{r+1,i} =B$ lest the two kings then attack one another. As a result, 25 of the subsequent layer is fixed as $B$, leaving the remaining $25$ to be fixed as $T$ as well. The proof for the wall is analogous. $\blacksquare$ Claim: $a_1 = a_2 = \cdots = a_{25}$ and $a_{26} = a_{27} = \cdots = a_{50}.$ Similarly for the walls. Proof. Note that if there exists $a_j = T$ and $a_{j+1} = B,$ then there can be no $b_i =L, b_{i+1} = R$ otherwise box $B_{i, j+1}$ and $B_{i+1,j}$ attack one another. Similarly, if $a_j = B$ and $a_{j+1} = T$ then there can be no $b_i = R, b_{i+1} = L.$ As a result, if there ever exists a sequence of $a_i \ne a_{i+1} = a_{i+2} \cdots = a_j \ne a_{j+1},$ then the corresponding labelling of the wall must be $b_1 = b_2 = \cdots = b_{50}.$ Contradiction. Hence, it must be that $a_i\ne a_{i+1}$ occurs only once, proving the claim. Same goes for $b_i.$ $\blacksquare$ In particular, the above result also proves that if $a_1 = a_2 = \cdots = a_{25} = T$ and $a_{26} = \cdots = a_{50} = B,$ then it must be true that $b_1 = b_2 = \cdots = R $ and $b_{26} = \cdots = b_{50} = L.$ Likewise, if $a_1 = a_2 = \cdots = a_{25} = B$ and $a_{26} = \cdots = a_{50} = T,$ then it must be true that $b_1 = b_2 = \cdots = L $ and $b_{26} = \cdots = b_{50} = R.$ Hence, there's only two possible arrangements.

This problem is basically Canada 2019/3

Here's a different solution: The answer is $\boxed{2}$. Note that if we combine kings of two consecutive rows and put them in row, then no two of those $50$ can be consecutive. Call a $50$ element subset of $\{1,2,\ldots,100\}$ sparse if no two elements are consectutive. Note that only sparse set containing $2$ is all even number and containing $99$ is odd numbers. Now consider the set $\mathcal R_2$ of all rows which are same as some row containing $2$ or is adjacent to such a row. Define $\mathcal R_{99}$ similarly. Because of our previous observations, we obtain any number in a row of $\mathcal R_2$ must be even, forcing $|\mathcal R_2| \le 50$. Similarly, $|\mathcal R_{99}| \le 50$. Moreover, $\mathcal R_2 \cap \mathcal R_{99} = \emptyset$. Note that usually $\mathcal R_2$ and $\mathcal R_{99}$ are big and for both these inequalities to hold, the two set of rows for $2,99$ must be $\{1,3,\ldots,49\},\{100,98,\ldots,52\}$ in some order. Lets assume $2$ has the first set. In this case we want to prove there is only one admissible confi. Note that the $50$th row has all even numbers and the $51$th row has all odd. Considering the union of these, we obtain a sparse set with exactly $25$ even and $25$ odd numbers, which does not contain both of $2,99$. Difference between a consecutive odd and even in this must be $2$ and there must be exactly one such consecutive pair, so we obtain that this sparse set must be $1,3,\ldots,49,52,54,\ldots,100$. So $50$th row contains $52,54,\ldots,100$ and the $51$th row contains $1,3,\ldots,49$. This will determine everything else uniquely. $\blacksquare$

First, we notice that if we divide the $100\times100$ chessboard into 2500 $2\times2$ blocks, each block cannot contain more than one king due to rule (i). However, as we need $2500$ kings, each block must have one king. Call a block a UL block, UR block, BL block, BR block if the king is on the topleft, topright, bottomleft, and bottomright corner of the block, respectively. Also, a UL block is also defined to be a U block and a L block, etc. Lemma: In every row of blocks, the blocks must all be a L block or all be a R block. There are $25$ of these L block rows(call them L rows) and $25$ of these R block rows(call them R rows). Proof: Consider all $50$ rows of blocks. On the leftmost two columns, there must be $25$ L blocks and $25$ R blocks to satisfy rule (ii). However, a R block automatically forces all blocks to the direct right of it to be an R block due to rule (i). Therefore, to satisfy rule (ii) on the other columns, the rest of the blocks must all be L blocks. Note that by symmetry, there are also $25$ U columns and $25$ D columns. Lemma: All L rows are consecutive, and all R rows are consecutive. Proof: FTSOC assume the above is not true. WLOG assume that not all L rows are consecutive. This means that there are two consecutive rows that go from L to R and two that go from R to L. It is clear that there are two consecutive columns that are one U and one D, so we can split into cases. If the two consecutive columns go from U to D, the L to R rows and the U to D columns intersect at four blocks that don't satisfy rule (i). If the two consecutive columns go from D to U, the same problem happens with the R to L rows. Therefore, we have reached a contradiction, so all L rows are consecutive, and all R rows are consecutive. By symmetry, all U columns are consecutive, and all D columns are consecutive. At this point, we have narrowed the number of possible boards to $2\cdot2=4$, because the kings' positions in the blocks are determined by the rows and columns, and there are only two possible arrangements for rows and two possible arrangements for columns. Upon further inspection on the four blocks in the center, only $\boxed{2}$ possible arrangements work due to rule (i): L to R rows, D to U columns and R to L rows, U to D columns. QED.

Divide the $100\times 100$ board into $2500$ $2\times 2$ boxes. Note that if there are two kings in one of these boxes, then they must attack each other. Thus, each of the boxes must contain exactly one king. $~$ Call each of these boxes dark if the king resides in one of its top squares and light if the king resides in one of its bottom squares. Call it blue if the king is on one of the left squares and red otherwise. $~$ Note that in each row of $50$ boxes, there must be $25$ dark ones and $25$ light ones. Furthermore, if a light box has a dark box below it, this is a contradiction. Therefore, for every box in the top row that is light, all the ones directly below it must also be light. Consequently, since that creates $25$ light boxes in each row, all other boxes must be light. $~$ Note that the analogous holds for columns: all boxes in the same column have the same hue, with $25$ red and $25$ blue, all boxes in the same row have the same color, with $25$ dark and $25$ light. Thus, we may assign each column with a hue and each row with a color and the box is determined by the intersection. $~$ Consider two neighboring columns which have different hues. This must exist, because there are both dark and light columns. Now, assume that there are two pairs of adjacent rows with different colors, one with blue above red, and one with red above blue. Now, through casework one sees that one of these must have two boxes sharing only a vertex such that their kings attack each other. $~$ Therefore, there is only one pair of adjacent rows with different colors, and analogously, only one pair of adjacent columns with different hues. We can only point out one place for which the color and the hues differ, because there are 25 of each color or hue. Furthermore, fixing the rows results in fixing the columns, because exactly one of the ways to distribute colors into the rows causes two kings to attack, as shown before. Thus, there are just two ways to place kings.

Solved with IshanR and Epicbird08. Partition the board into $2500$ 2x2 squares, and observe that each square clearly must contain exactly one king. Define a square's string to be LU if the king is in the top left, RU if it is in the top right, LD if it is in the bottom left, and RD otherwise. Define a square to be L if it has L in its string, R if it has R in its string, etc. We now claim that if some square has R in its string, then all squares directly to its right have R. This is obvious, and works for the other three directions. Observe that the leftmost column of the 2x2 squares must have $25$ squares with L, and $25$ squares with R, and for each of the latter $25$ squares all squares to the right are also R. This shows that there are at least $25 \cdot 50 = 1250$ squares with R. However, if we rotate the large 100x100 square board 180 degrees and apply the. same argument, we obtain that there are at least $1250$ squares with L. Thus, there are $1250$ squares of both types, and in particular, each row is either completely L or completely R. This also holds for the columns. Now, instead of considering labelings of individual 2x2 squares, consider labelings of the rows of 2x2 squares and the columns. Each row can clearly be labeled with either L or R, and each column can be labeled with U or D. We now claim that if there is a R above an L, there cannot be a D to the left of a U. This is obvious, because otherwise the intersection of these two rows and columns gives a contradiction. Additionally, if there is a R below an L, there cannot be a D to the right of a U for the same reason. Thus, there clearly cannot be a R both above and below an L, so the final arrangement of Rs and Ls can only consist of a string of Rs next to a string of Ls, with the same holding for Us and Ds. We finally get two configurations, one with the rows (from top to bottom) being RRRR...LLLL.... and the columns (from left to right) being UUUU....DDDD...., and one with the rows being LLLL....RRRR.... and the columns being DDDD....UUUU...., and the proof is complete.

Divide the board into $2 \times 2$ squares. Two kings in the same square will attack each other, so we must have exactly one king in each square. Call a square $U$ if the king is in the top row, and $D$ if the king is in the bottom row. If there is a $U$ square below a $D$ square, then the kings will attack each other, so in each column of squares there will only be one of $U$ or $D$ squares. Call a square $R$ if the king is on the right, and $L$ if the king is on the left. We can similarly show that each row contains only $R$ squares or only $L$ squares. If a $U$ column is to the right of a $D$ column, then there cannot exist a $R$ row above an $L$ row, otherwise the kings will attack each other. Similarly, if there is a $D$ column to the right of a $U$ column, then there cannot exist an $L$ row above an $R$ row. Therefore, all the $U$ columns and all the $D$ columns must be next to each other, and similarly, all the $R$ columns and all the $L$ columns must be next to each other. The placing of the columns will determine the placing of the rows. Thus, there are $2$ arrangements.

Divide the board into $50^2=2500$ $2 \times 2$ squares. In each there is not more than one king, so actually exactly one king Write in the square R if the king is in the right column of it, and L if in the left. Consider any two columns of squares (so subboard $4 \times 100$). If in the left column some square has R in it, then in the right column the coresponding square must also have R in it. Because in each column we have $25$ kings, it must also be true with L's. Hence, in every row of squares all $50$ squares have the same letter written in it. Write this letter near the row Same holds, if we write U and D near columns Now to avoid getting two kings capturing diagonally, we cannot have UD near columns and RL near rows, and DU near columns and LR near rows Note that rows must have one of LR or RL, so columns cannot have both UD and DU, thus it is either $U\ldots UD\ldots D$, then $L\ldots LR\ldots R$, or $D\ldots DU\ldots U$, then $R\ldots RL\ldots L$. So only two possible arrangements