We look at the suits as four different colors red, blue, green, and magenta. We will be denoting cards by bullets (i.e. $\bullet$). Observe the following things: If two cells of the same suit are diagonally placed, then there two common adjacent cells must also be of the same suit. In particular, if we have some configuration like $$ \begin{matrix} \bullet & \color{red}{\bullet} \\ \color{red}{\bullet} & \bullet \end{matrix} $$Then both the $\bullet$ must have color red. If we some configuration like $$\begin{matrix} \color{blue}{\bullet} & \bullet \\ \color{red}{\bullet} & \color{red}{\bullet} \end{matrix}$$Then the $\bullet$ must have color blue. This gives that any connected component of a particular color must be a rectangle (using first point), and moreover, all rectangles thus obtained must be aligned (using second point). For example, the first figure below shows not aligned rectangles while second shows aligned rectangles: $$ \begin{matrix} \color{blue}{\bullet} & \color{blue}{\bullet} & \\ \color{blue}{\bullet} & \color{blue}{\bullet} & \\ \color{blue}{\bullet} & \color{blue}{\bullet} & \\ \color{red}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} \\ \color{red}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} \end{matrix} \qquad \qquad \begin{matrix} \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} \\ \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} \\ \color{blue}{\bullet} & \color{blue}{\bullet} & \color{blue}{\bullet} \\ \color{red}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} \\ \color{red}{\bullet} & \color{red}{\bullet} & \color{red}{\bullet} \end{matrix} $$Now we will be denoting a rectangle by $\mathcal R$. Note that any two rectangles of the same color cannot be adjacent nor be diagonally placed (otherwise we obtain bigger rectangles). So we have some representation like $$ \begin{matrix} \mathcal R & \cdots & \mathcal R \\ \vdots & & \vdots \\ \mathcal R & \cdots & \mathcal R \end{matrix} $$ Claim: If some three rectangles of some column have configuration like (there possible can be one more rectangle in the column) $$ \begin{matrix} \color{red}{\mathcal R} \\ \color{green}{\mathcal R} \\ \color{blue}{\mathcal R} \end{matrix} $$Then that column of rectangles must be the only one (i.e. there are no more columns of rectangles). Proof: If not, then somewhere we must obtain some configuration like following : $$ \begin{matrix} \color{red}{\bullet} & \color{blue}{\bullet} \\ \color{green}{\bullet} & \color{magenta}{\bullet} \\ \color{blue}{\bullet} & \color{red}{\bullet} \end{matrix} $$But note that we cannot assign numbers here without a contradiction. $\square$ Claim: (Key Claim) Rectangles of the first column contain all four colors. Proof: Assume contrary. Now lets consider how many colors do the rectangles of the first column contain. $1$ color. Then each color must occur a multiple of $4$ times. We directly get a contradiction as $4 \nmid 13$ $2$ colors, say say $\mathcal C_1,\mathcal C_2$. Then it is not hard to note that second row must contain rectangles of $\mathcal C_3, \mathcal C_4$ only, third row contain $\mathcal C_1,\mathcal C_2$ only and so on. Three examples are given below: $$ \begin{matrix} \color{red}{\mathcal R} & \color{magenta}{\mathcal R} & \color{red}{\mathcal R} & \color{green}{\mathcal R} & \color{blue}{\mathcal R} & \cdots \\ \color{blue}{\mathcal R} & \color{green}{\mathcal R} & \color{blue}{\mathcal R} & \color{magenta}{\mathcal R} & \color{red}{\mathcal R} & \cdots \end{matrix} \qquad \begin{matrix} \color{red}{\mathcal R} & \color{blue}{\mathcal R} & \color{red}{\mathcal R} & \color{blue}{\mathcal R} & \cdots \\ \color{green}{\mathcal R} & \color{magenta}{\mathcal R} & \color{green}{\mathcal R} & \color{magenta}{\mathcal R} & \cdots \\ \color{red}{\mathcal R} & \color{blue}{\mathcal R} & \color{red}{\mathcal R} & \color{blue}{\mathcal R} & \cdots \end{matrix} \qquad \begin{matrix} \color{red}{\mathcal R} & \color{green}{\mathcal R} & \color{blue}{\mathcal R} & \color{magenta}{\mathcal R} & \cdots \\ \color{blue}{\mathcal R} & \color{magenta}{\mathcal R} & \color{red}{\mathcal R} & \color{green}{\mathcal R} & \cdots \\ \color{red}{\mathcal R} & \color{green}{\mathcal R} & \color{blue}{\mathcal R} & \color{magenta}{\mathcal R} & \cdots \\ \color{blue}{\mathcal R} & \color{magenta}{\mathcal R} & \color{red}{\mathcal R} & \color{green}{\mathcal R} & \cdots \\ \end{matrix} $$That implies number of color $\mathcal C_1$ plus of $\mathcal C_2$ in the whole grid is divisible by $4$, contradiction (as $4 \nmid 26$). $3$ colors. Because of the previous claim, the configuration of first column must look something like $$ \begin{matrix} \color{red}{\mathcal R} \\ \color{green}{\mathcal R} \\ \color{blue}{\mathcal R} \end{matrix} $$Then the consequent columns must look like the following: $$ \begin{matrix} \color{red}{\mathcal R} & \color{blue}{\mathcal R} & \color{red}{\mathcal R} & \color{blue}{\mathcal R} & \cdots \\ \color{green}{\mathcal R} & \color{magenta}{\mathcal R} & \color{green}{\mathcal R} & \color{magenta}{\mathcal R} & \cdots \\ \color{blue}{\mathcal R} & \color{red}{\mathcal R} & \color{blue}{\mathcal R} & \color{red}{\mathcal R} & \cdots \end{matrix} $$Now we look at the colors green, magenta. As they are color of $26$ things, so this means the middle rectangles of all rows have width $2$. But in that case if we look at only one of the colors (green or magenta), then we get a contradiction as $2 \nmid 13$. This proves our claim. $\square$ Now the two claims just finish the problem. $\blacksquare$ Remark: If $13$ is replaced by any odd number, then a similar proof as above works. But if $13$ is replaced by some even number, then the assertion of the problem is not true. For example: If $13$ is replaced by $4$: $$ \begin{matrix} \color{blue}{4} & \color{blue}{3} & \color{green}{3} & \color{green}{4} \\ \color{blue}{2} & \color{blue}{1} & \color{green}{1} & \color{green}{2} \\ \color{red}{2} & \color{red}{1} & \color{magenta}{1} & \color{magenta}{1} \\ \color{red}{4} & \color{red}{3} & \color{magenta}{3} & \color{magenta}{4} \end{matrix} $$ If $13$ is replaced $6$: $$ \begin{matrix} \color{green}{6} & \color{green}{5} & \color{green}{4} & \color{red}{4} & \color{red}{5} & \color{red}{6} \\ \color{blue}{6} & \color{blue}{5} & \color{blue}{4} & \color{magenta}{4} & \color{magenta}{5} & \color{magenta}{6} \\ \color{blue}{1} & \color{blue}{2} & \color{blue}{3} & \color{magenta}{3} & \color{magenta}{2} & \color{magenta}{1} \\ \color{red}{1} & \color{red}{2} & \color{red}{3} & \color{green}{3} & \color{green}{2} & \color{green}{1} \end{matrix} $$ It is not hard to generalize these for any $2k$, ($k \ge 2$).

hard to explain without diagrams Throughout this solution "adjacent" means vertical/horizontal by default and "diagonally adjacent" will be used when referring to the obvious interpretation. View the suits as four different colors $a,b,c,d$. Note the following facts: If two cells are diagonally adjacent and have the same color, then the two cells adjacent to both are also that color No cell can be adjacent to (at least) three different-colored cells, since then we get a contradiction by trying to fill in the rest of the $2 \times 3$ rectangle containing all four cells. In particular the ends of $1 \times n$ or $n \times 1$ rectangles must share an edge with the $13 \times 4$ grid If a cell is adjacent to two different-colored cells, then the cell adjacent to both these two cells (also diagonally adjacent to the original) must be the fourth color. That is, we cannot have three colors "meeting" at a point The first fact alone implies that every "corner-connected" component is also "edge-connected", and that every connected component is a rectangle (place a rectangle in the smallest rectangle containing it, and if there is some connected components of "unfilled" cells we can look along its border and get a contradiction). Now the idea is that the entire grid must be tiled with the same "rectangle pattern" like shown below: [asy][asy] draw((0,0)--(13,0)--(13,4)--(0,4)--cycle); draw((0,1)--(13,1)); draw((0,3)--(13,3)); draw((4,0)--(4,4)); draw((5,0)--(5,4)); draw((7,0)--(7,4)); draw((11,0)--(11,4)); [/asy][/asy] To see this we induct starting from the leftmost edge. Call a group of rectangles aligned if their left and right edges, when extended to lines, are all shared. Then it's easy to see the rectangles sharing the leftmost edge with the big $13 \times 4$ grid must all be aligned. Then the horizontal edges in this first group of rectangles must "extend"—check that we can't lose any, and that if we add a new horizontal edge this contradicts the third bullet point. Repeating this argument until we reach the rightmost edge finishes. Now perform casework on the "rectangle pattern" (essentially listing how tall they are in order). Deciphering the meaning of this passage is left as an exercise Pattern is $4$. Then every suit must have a number of cards divisible by $4$: absurd Pattern is $3,1$ or $2,2$ or $1,3$. Then if the leftmost rectangles are $a,b$ in some order the next rectangles are $c,d$ in some order, and then the next rectangles are $a,b$ in some order, and so on. Thus the total number of $a$ and $b$ cards (together) will be divisible by $4$, but this is not the case Pattern is $1,2,1$. Then if the leftmost rectangles are $a,b,c$ in that order the next set of rectangles are $c,d,a$ in that order, and then $a,b,c$ again, and so on. But then the number of $b$ cards is divisible by $2$ Pattern is $1,1,2$ or $2,1,1$: this contradicts the second bullet point unless the middle $1$-tall rectangle touches both the leftmost and rightmost edge, but this would mean (by looking at the $2$-tall rectangle) some suit has $26$ cards Pattern is $1,1,1,1$. For the same reason as above these rectangles have to touch both the leftmost and rightmost edge, which implies the desired result We are done. $\blacksquare$