$2 \times$ sum of all numbers $= (1+2+...+2n)=n(2n+1)$ and so $2|n$ If $n=2k$ then exists such way: Row $i$ where $1 \leq i \leq k$ we fill with $i$ numbers $1$ in the beginning of row and other are $0$. Row $i$ where $k+1 \leq i \leq 2k$ we fill with $2k-i$ numbers $1$ in the end of the row and other are $2$. As example for $n=8$ $10000000$ $11000000$ $11100000$ $11110000$ $22222111$ $22222211$ $22222221$ $22222222$ Easy to show, that rows contain sums $1,2,...,k$ and $3k+1,3k+2,...,4k$ and columns contain sums $k+1, k+2,....,3k$

Note that every cell contributes to two of the sums $1, 2, \ldots, 2n$ (once on its row and once on its column). So the sum $1 + 2 + \dots + 2n = n(2n + 1)$ must be even, which in turn means $n$ is even. For even $n$ we can construct such a table by induction. For $n=2$ observe that table with rows $(0, 2), (1, 2)$ satisfies the problem. Let $X_n$ be a table with dimensions $n\times n$ whose entries are $0$, $1$ or $2$ and column and row sums are $1, 2, \ldots, 2n$. Consider table $$\begin{matrix} &&& & 0 & 2\\ &&X_n& & 0 & 2\\ &&& & 0 & 2\\ 0& 0&\ldots &0 & 0 & 2\\ 2& 2&\ldots & 2 & 1&2\\ \end{matrix}$$ which has dimensions $(n+2) \times (n+2)$. This new table extends every column and row sum of $X$ by $2$ while the remaining $4$ new rows and columns give us sums $1, 2, 2n+3, 2n+ 4$, Therefore the new board has row and column sums $1, 2, \ldots, 2n+3, 2n+4$. This finishes our induction.

Given that every cell contributes to one row and one column, then it contributes twice on the sum $1, 2, \ldots, 2n$. So the sum $1 + 2 + \dots + 2n = n(2n + 1)$ must be even, which in turn means $n$ is even. $n=2k$ Checking the first 3 even values of $n$: $2,4,6$ We can observe a beautiful pattern made by the original $2\times 2$ yellow block, blocks of 0's (in red) and blocks of 2's (in blue). So basically we see that for every value of $k$, we can construct a table of $k\times k$ blocks, where the diagonal is made by yellow blocks, the upper left is filled by red blocks and the lower right by blue blocks. Quote: Note that the original $2\times 2$ contains rows with sums $\equiv 1,0 \pmod{4}$ and columns with sums $\equiv 2,3 \pmod{4}$ so: when adding a blue block, each row/column increases by 4, and adding a red block doesn't change the sum, so we're maintaining the congruences for row sums and column sums $\mod{4}$ For any given $n\times n$ table with this block construction, row sums end in $2n-3$ and $2n$, while column sums end in $2n-2$ and $2n-1$. Now, we can construct the $n+2\times n+2$ table by overlapping the same $n\times n$ table as the second image shows. Note that previous row sums and column sums are preserved because of the red blocks. So all we need to do is add the final blue block that adds 4 to each of the last 2 rows and column sums, therefore creating the extra 4 values $2n+1,2n+2,2n+3,2n+4$.

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