a) For $n = 4$ the filling below is good.

b) We prove that there is no good filling of an $n \times n$ square if $n$ is odd. Assume the contrary to be true. Then, for some odd number $n$, we have that, for every index $i$, $1 \le i \le n$, the row no. $i$ and the column no. $i$ contain, together, exactly once, each element of $M$. If we denote by $a_{ij}$ the number situated on row no. $i$ and column no. $j$, $i, j \in \{1, 2, \dots, n\}$, and $S_i = a_{i1} + a_{i2} + \dots + a_{in} + a_{1i} + a_{2i} + \dots + a_{ni}$ in which the term $a_{ii}$ only appears once, then each element of $M$ must appear in $n$ such sums. But if $a_{ij} = k, i \neq j$, then the element $k$ will contribute to both $S_i$ and $S_j$. Thus, any number that is not situated on the main diagonal contributes to two sums. But in total, there is an odd number of sums, so each element of $M$ must appear an odd number of times on the main diagonal. Thus, each element of $M$ needs to appear at least once on the main diagonal, which is not possible because there are $2n-1$ elements in $M$ and only $n$ places on the diagonal.