Okay, this one is cute. (a): If the two row sums are already distinct, then we have nothing to care about; so suppose they are equal. If there is a column with two distinct numbers, then swapping them would fix the issue. Otherwise, say each column contains a pair of the same number. So the table looks like this:$$\begin{array}{|c|c|c|c}\hline a_1 & a_2 & a_3 & \quad \cdots\quad \\ \hline a_1 & a_2 & a_3 & \cdots \\ \hline \end{array}$$Suppose WLOG that $a_1,a_2,a_3$ are the biggest three elements. Since the column sums $2a_1,2a_2,\cdots ,2a_n$ are all distinct, we have that these $a_i$'s are all distinct. Consider a new, tweaked array:$$\begin{array}{|c|c|c|c}\hline a_1 & a_3 & a_3 & \quad \cdots\quad \\ \hline a_2 & a_2 & a_1 & \cdots \\ \hline \end{array}$$Note that the column sums are still different; because if we had, say $a_1+a_2=a_3+a_2$, then we'd get $a_1=a_3$, which is false. Also, we can't have, say, $a_1+a_2=a_i+a_i$ for some $i\not\in\{1,2,3\}$, because $a_1,a_2>a_i$. Now the row sums must have become distinct, or we'd have $a_1+a_3+a_3=a_2+a_2+a_1\implies a_2=a_3$, which is wrong. This finishes (a). $\blacksquare$ (b): We claim that such a permutation isn't always feasible. Before we get to the proof, we need a small lemma: Lemma: Suppose $n\ge 3$ is an integer. If $x_1,x_2,\cdots ,x_n$ are distinct nonegative integers satisfying $$x_1+x_2+\cdots +x_n=\frac{n^2-n+1}{2},$$then $(x_1,x_2,\cdots ,x_n)$ is $(0,1,2,\cdots ,n-3,n-2,n)$ upto permutation. Proof: We induct on $n$. The base case $n=3$ is obvious (and the $n=4$ case is well-known to Kakuro solvers); let's prove it for general $n>3$. If all of $x_i$'s were nonzero, then we'd have $$x_1+x_2+\cdots +x_n\ge 1+2+3+\cdots +n=\frac{n^2+n}{2}>\frac{n^2-n+2}{2}.$$Of course, this is bad. Therefore one of the $x_i$'s is $0$; suppose WLOG $x_1=0$. Then the $n-1$-tuple $(x_2-1,x_3-1,\cdots ,x_n-1)$ satisfies the conditions of the induction hypothesis, and this gets our job done. $\square$ Now for the main proof. Consider the $100\times 100$ array defined as follows: fill the second column with $1$ one, third column with $2$ ones, fourth column with $3$ ones, $\cdots$, $99-$th column with $98$ ones. The $100-$th column, is however, filled with $100$ ones. The remaining cells are all stuffed with zeros. Of course, the column sums are all different. Suppose there's a way to permute these $4951$ ones so that column sums are distinct and the row sums are distinct as well. Suppose $x_i$ is the number of ones in the $i-$th column and $y_i$ is the number of ones in the $i-$th row. Note that $x_1,\cdots ,x_{100}$ are distinct nonnegative integers summing to $4951=\tfrac{100^2-100+2}{2}$, so our lemma says that $(x_1,\cdots ,x_{100})$ is a permutation of $(0,1,\cdots , 98,100)$. Suppose $x_a=100$. In a similar manner, we see that $(y_1,\cdots ,y_{100})$ is a permutation of $(0,1,\cdots , 98,100)$; say $y_b=0$. Now the $a-$th column has all $100$ cells filled with ones, and $b-$th row has no one at all, so there's a Schrödinger's one at their intersection, which is not allowed in combinatorics. Done! $\blacksquare$