We claim that this minimum is $ m = 4(n - k)$. This is achievable by placing pawns along a few diagonals which are hard to describe without a proper diagram. To see that this is the minimum, we divide the board up as follows: [asy][asy]size(200); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); dot((0.4,1)); dot((0.6,1)); dot((1,0.4)); dot((1,0.6)); draw((0.4,1)--(0.4,0)); draw((0.6,1)--(0.6,0)); draw((1,0.4)--(0,0.4)); draw((1,0.6)--(0,0.6)); label("$2k-n$",(0.5,1),N); label("$n-k$",(0.2,1),N); label("$n-k$",(0.8,1),N); label("$2k-n$",(1,0.5),E); label("$n-k$",(1,0.2),E); label("$n-k$",(1,0.8),E);[/asy][/asy] Consider each oblong rectangular subfigure. Each empty row/column in these (whichever is perpendicular to the longer sides) necessitates two pawns; one in each of the adjoining $ n - k$ x $ n - k$ squares. We then divide by two because we go over each square block at most twice, and add $ \sum_{cyc} n - k -$(number of empty rows/columns in each rectangular box) to account for the pawns which are in the rectangular blocks themselves to get $ 4(n - k)$ pawns - everything else cancels out. Motivation: Divide each side into $ k$ and $ n-k$ blocks from the left and the right (pretty obvious approach), then notice that we can use the rectangular boxes to count corners. It would seem intuitively that placing pawns along the diagonal of the central box would be less "wasteful" - anyone have a $ 4(n - k)$ example/solution with that? I actually tried that approach for quite a while, but it seemed fruitless to me.

Actually, I think $m = 3n - 2k$. I have a model for this and a solution. I am still searching my mistake in my proof because the answer should imply that $k \le \dfrac{2}{3}n$. Anyone have other results ?(could there exist a typo and instead of $\dfrac{2}{3}$ should be $\dfrac{3}{2}$ ?)

Now fwolith's proof is correct.Your assertion is not valid for $n=6$(I had found a coloring of 9 squares,while your answer is 10)

My solution: Number the rows and columns from $0$ to $n-1$.Color every cell by one of the numbers $0$ to $k-1$ such that $i+j+1 \equiv i'+j'+1 \pmod{k}$ means $(i,j)$ and $(i',j')$ have the same number,$(0,0)$ contains $1$,$(0,1)$ contains two,....,$(0,k-1)$ contains $k$,$(0,k)$ contains $0$ again and so on.Now place the pawns on the cells numbered $0$.By the inequality $\frac{n}{2} < k \le \frac{2n}{3}$,$(i,j)$ is coloured $0$ iff and only if $i+j+1=k,2k$ or $3k$.$i+j+1=k$ has $k$ solutions:$(0,k-1),(1,k-2),\cdots,(k-1,0)$.$i+j+1=2k$ has $2n-2k$ solutions:$(2k-n,n-1),(2k-n+1,n-2),\cdots,(n-1,2k-n)$.$i+j+1=3k$ has $2n-3k$ solutions:$(3k-n,n-1),(3k-n+1,n-2),(n-1,3k-n)$.So the total no of cells containing $0$ by this coloring is $k+(2n-2k)+(2n-3k)=4(n-k)$. Now we show that this is the minimum.Every $k$ consecutive cells should have atleast one pawn.Now suppose that in the coloring mentioned above all pawns are not placed over cells having same number.Then there becomes a gap of $k$ consecutive cells (considering that we place one pawn in each k-length block in order to achieve the minimal coloring).So every pawn should be placed over cells having same number.It is easy to verify that the no of $0$'s are the least,so this is the required minimal coloring.

Here is nice solution. First make coordinate system with lower left square being (1,1) (next on right (2,1),one up from (1,1) is (1,2) and so on..) Put pawns on all squares with sum of coordinates k+1,2k+1,3k+1 and (1,1) if it is possible.(k+1 and 2k+1 are always possible and 3k+1 is impossible if k=[(2/3)*n]).There are 4(n-k) pawns and it is easily checked that there are no k consecutie empty squares .To prove that you need at least 4(n-k) pawns we can make four k(n-k) rectangles and fit them inside big square(construction is well known but idk how to describe it without picture).In every rectangle we need n-k pawns so we need at least 4(n-k) pawns.Only problem with first contruction is if k=4L where n=6L , we get 4(n-k)+1 .But we see if we remove pawn from (1,1) it is still satisfying condition and we are done.(this is just a fency way of summing 1,5 hour of work)

I claim that the answer is $m=4(n-k)$. To show that it is sufficient, consider the following diagram (thanks, fwolth): [asy][asy]size(200); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); dot((0.4,1)); dot((0.6,1)); dot((1,0.4)); dot((1,0.6)); draw((0.4,1)--(0.4,0)); draw((0.6,1)--(0.6,0)); draw((1,0.4)--(0,0.4)); draw((1,0.6)--(0,0.6)); label("$2k-n$",(0.5,1),N); label("$n-k$",(0.2,1),N); label("$n-k$",(0.8,1),N); label("$2k-n$",(1,0.5),E); label("$n-k$",(1,0.2),E); label("$n-k$",(1,0.8),E); label("$A$", (0.15,0.2),E); label("$B$",(0.5,0.15),N); [/asy][/asy] Consider only the outer $n-k$ rows/columns on each side. Also, define the weight of a pawn within a row/column to be $1/2$ if it lies within a region like $A$, and $1$ if it lies within a region like $B$. Then, in order to satisfy the condition, each row/column needs to have total weight at least 1, namely either one pawn in its intersection with region $B$, or 1 in each of the intersections with region $A$. Now, each pawn contributes to the total weight by 1, since pawns in region $A$s are counted for both a row and column, and we wish to have net weight $4(n-k)$, for all of the rows/columns. Therefore, we need at least $4(n-k)$ pawns. To show this is sufficient, consider the construction below: [asy][asy]size(200); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); dot((0.4,1)); dot((0.6,1)); dot((1,0.4)); dot((1,0.6)); draw((0.4,1)--(0.4,0)); draw((0.6,1)--(0.6,0)); draw((1,0.4)--(0,0.4)); draw((1,0.6)--(0,0.6)); draw((0.4,0.8)--(0.6,1),red); draw((0.8,0.6)--(1,0.8),red); draw((0.2,0.6)--(0.4,0.8),red); draw((0,0.4)--(0.2,0.6),red); draw((0.2,0)--(0.4,0.2),red); draw((0.4,0.2)--(0.6,0.4),red); draw((0.8,0)--(1,0.2),red); draw((0.6,0.4)--(0.8,0.6),red); label("$2k-n$",(0.5,1),N); label("$n-k$",(0.2,1),N); label("$n-k$",(0.8,1),N); label("$2k-n$",(1,0.5),E); label("$n-k$",(1,0.2),E); label("$n-k$",(1,0.8),E); [/asy][/asy]

The answer is $4(n-k)$. Mark each point on the chessboard with a coordinate $(x,y)$ where the bottom left square is $(0,0)$, the square above $(i,j)$ is $(i,j+1)$ and the square to the right of $(i,j)$ is $(i+1,j)$. Place a pawn each square $(i,j)$ for which $k\mid i+j$. Suppose there exist a line of $k$ consecutive unoccupied squares. Without loss of generality, let them be in the same row and let them be $(i,j+r)$ for $0\le r\le k-1$. However, let $r\equiv -i-j\pmod k$ then $k\mid i+j+r$, contradiction. To prove that at least $4(n-k)$ is necessary, call a pawn row-strong if its $x$-coordinate is between $n-k$ and $k-1$, inclusive. Define column-strong similarly. If a row does not have a row-strong pawn, then there needs to be at least two pawns in that row. There are $2(n-k)$ columns which make a pawn not row-strong, and each of these columns must have a pawn. Call a row or column whose fixed coordinate is not between $n-k$ and $k-1$ inclusive strong if it has a row-strong pawn and a column-strong pawn, respectively. Let there be $r_0$ strong rows and $c_0$ strong columns. There are $2n-k$ rows/columns not between $n-k$ and $k-1$. Each non-strong row or non-strong column must have two pawns. Thus, there are at least \[\text{max}(2(2(n-k)-r_0), 2(2(n-k)-c_0))\ge 4(n-k)-r_0-c_0\]pawns in non-strong rows. These pawns are neither row-strong nor column-strong. There are at least $r_0$ row-strong pawns and $c_0$ column-strong pawns. Thus, there are at least $4(n-k)$ pawns in total.

A simplified version of the proof $m\ge 4(n-k)$ first divide each side to three sections $n-k,\ 2k-n,\ n-k$ dividing the board into 9 as shown in the diagram (tnx, fwolth): [asy][asy]size(200); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); dot((0.4,1)); dot((0.6,1)); dot((1,0.4)); dot((1,0.6)); draw((0.4,1)--(0.4,0)); draw((0.6,1)--(0.6,0)); draw((1,0.4)--(0,0.4)); draw((1,0.6)--(0,0.6)); label("$2k-n$",(0.5,1),N); label("$n-k$",(0.2,1),N); label("$n-k$",(0.8,1),N); label("$2k-n$",(1,0.5),E); label("$n-k$",(1,0.2),E); label("$n-k$",(1,0.8),E);[/asy][/asy] Consider the upper $2k-n\times n-k$ rectangle containing $n-k$ rows. if one of these rows does not have a pawn inside of this rectangle, then it must have at least 2 pawns, one inside each $n-k\times n-k$ square on the sides. Otherwise there will be $k$ consecutive squares not containing a pawn. Now assume $a_1$ rows of the upper $n-k$ rows do not have a pawn inside the $2k-n\times n-k$ rectangle. Similarly $a_2$ rows from the bottom and $a_3,\ a_4$ columns from the sides do not have any pawns in the middle rectangle. Thus the number of pawns should be at least $(n-k-a_1)+(n-k-a_2)+(n-k-a_3)+(n-k-a_4)+(2a_1+2a_2+2a_3+2a_4)/2=4(n-k)$ note: divide the $a_i$'s by two because each pawn in the middle will be counted at most twice