No. Let the number of stones
in the three piles be \(a, b\) and \(c\),
respectively. Consider (mod \(3\)) of these
numbers. In the beginning, they are \(1, 2, 0\). After one operation, they become \(0, 1, 2\) no matter which two piles have stones
transfer to the third pile. So the
remainders are always \(0, 1, 2\) in some
order. Therefore, all piles having \(12 \) stones are impossible. 

it's very easy. you should do it yourself

 Let \(S\) be the sum of all the \(mn\) numbers in the table. Note that after an operation, each number stay the same or turns to its negative. Hence there are at most \(2^{mn}\) tables. So \(S\) can only have finitely many possible values. To make the sum of the numbers in each line nonnegative, just look for a line whose numbers have a negative sum. If no such line exists, then we are done. Otherwise, reverse the sign of all the numbers in the line. Then \(S\) increases. Since \(S\) has finitely many possible values, \(S\) can increase finitely many times. So eventually the sum of the numbers in every line must be nonnegative. 

In the beginning randomly pair the points and join the segments. Let \(S \) be the sum of the lengths of the segments. (Note that since there are finitely many ways of connecting \(2n \) points by \(n \) segments, there are finitely many possible values of \(S\).) If two segments \(AB\)and \(CD\)intersect at \(O\), then replace pairs \(AB\)and \(CD\)by \(AC\)and \(BD\). Since

\(AB + CD = AO + OB + CO + OD > AC + BD\)

by the triangle inequality, whenever there is an intersection, doing this replacement will always decrease \(S\). Since there are only finitely many possible values of \(S\), so eventually there will not be any intersection.

What did you try ?