$k=2681$. Proof of $k \geq 2681$: Abel must change at least 4021 of the sums, w.l.o.g. all column sums and all but at most one row sum. To change the column sums, Abel must alter at least one entry in each column. This Abel does w.l.o.g. first and then pauses. At the time of this pause, let $n$ rows be unaltered. Thus, Abel must alter an entry in $n-1$ of these rows so that the row sums can differ. This Abel does w.l.o.g. subsequently and pauses again. For $n \geq 671$, at least 2681 entries have already been altered. Now let $n \leq 670$. A cell is called insufficient if Abel has changed its entry but no further entry in the same row or column. Since the row and column sums of an insufficient cell are the same, no cell can be insufficient at the end. There are at least 2011 - $2 n$ insufficient cells during Abel's first pause. With each change up to the second pause, Abel can eliminate at most one of these insufficient cells (by changing an entry in the same column), and with each change after the second pause, at most two insufficient cells. Therefore, after the second pause, at least $\frac{1}{2}(2011-2 n-(n-1))$ entries must be altered, so in total at least $2011+(n-1)+\frac{1}{2}(2011-2 n-(n-1))=3016-\frac{n}{2} \geq 2681$.

Proof idea for $k \leq 2681$: For $1 \leq m, n \leq 2011$, let the cell in row $m$ and column $n$ be denoted by $\langle m ; n\rangle$. Abel increases the entry of 2681 cells $\langle m ; n\rangle$ each by $2(m+n)+1$, specifically from $\langle 2 ; 1\rangle$ and for $l=1, \ldots, 670$ from $\langle 3 l-1 ; 3 l-1\rangle,\langle 3 l ; 3 l-1\rangle,\langle 3 l+1 ; 3 l\rangle$ and $\langle 3 l+1 ; 3 l+1\rangle$.

Remarks: Sometimes it was argued that by changing two entries in the same row (or column), Abel can change three sums and this is optimal, and from this $k \geq 2681$ was concluded. However, it was implicitly assumed that an arrangement of cells with altered entries as in the proof of $k \leq 2681$ is already optimal.
Determine the smallest positive number $k$ for which the following is true: If Cain writes integers into the cells of a $2011 \times 2011$ chessboard in such a way that the 4022 sums obtained by adding all the numbers in a row or column are pairwise equal, then it is possible for Abel to achieve that these 4022 sums are pairwise distinct by altering the entries in only $k$ of the cells.