This is essentially a linear algebra problem. Let $z_i$ be the sum of the numbers in the $i$-th row (from top to bottom) and $s_i$ be the sum of the numbers in the $i$-th column (from right to left!) so that $s_n=1$. The condition now simply becomes that the number at position $(i,j)$ should be $s_i+z_j$. Hence we obtain the conditions $s_j=\sum_{i \le n+1-j} (z_i+s_j)$ and hence $\sum_{i \le n+1-j} z_i=(j-n)s_j$ and similarly $\sum_{i \le n+1-j} s_i=(j-n)z_j$. But this implies $z_1=s_1=0$ and $z_{j+2}=js_{n-j}-(j+1)s_{n-j-1}$ for $j=0,1,\dots,n-2$. Similarly, $s_{j+2}=jz_{n-j}-(j+1)z_{n-j-1}$. Substituting this into each other, we find that \[ z_{j}=j((n-j)z_{j}-(n-j+1)z_{j-1})-(j-1)((n-j-1)z_{j+1}-(n-j)z_{j})\]for $j=2,3,\dots,n-1$. For $j=2$, this implies that $z_3=z_2$ (unless $n<3$) and then inductively, we get that $z_{n-1}=z_{n-2}=\dots=z_2=z$. Then also $s_{n-1}=s_{n-2}=\dots=s_2=s=-z$. Finally, $s_n=(n-2)z$ and $z_n=(n-2)s$. Hence $z=\frac{1}{n-2}$ and $s=-\frac{1}{n-2}$. In other words, the only colouring is the one where the entries in the corners are $1,0,-1$, the entries along the boundary in between are $\frac{1}{n-2}$ on the one side and $-\frac{1}{n-2}$ on the other side, and all entries in the interior are $0$ (and this indeed works). As a bonus, we have proved that up to constant multiples, these are really all colourings (even without the condition that there is a $1$ in one corner).