Brazilian Math Olympiad

a. For instance,

1	2	3
4	5	6
8	9	7

The trick is to only adjust the last row. The usual order $7$, $8$, $9$ yields all sums to be multiple of $3$, so it's just a matter of rearranging them.

b. No, it's not possible. First, notice that the sum of three numbers $x$, $y$, $z$ is a multiple of $3$ iff $x\equiv y\equiv z(\mathrm{mod}3)$ or $x$, $y$, $z$ are $0$, $1$, $2$ mod $3$ in some order. Let $a$, $b$, $c$, $d$ be the numbers in the corner modulo $3$. So two of them are equal. We can suppose wlog that they are either $a=b$ or $a=d$. Also, let $x$ be the number in the central cell modulo $3$.

a	b
x
c	d

If $a=d$, then $x\ne a$ and $x$ is equal to either $b$ or $c$. Suppose wlog $x=b\ne a$. Then we have the following situation:

a	b
	b
c	a

Let $m$ be the other remainder (that is, $m\ne a$ and $m\ne b$). Then $m$ cannot be in the same line as $a$ and $b$. This leaves only one possibility:

a	b
m	b
m	m	a

But the remaining $a$ will necessarily yield a line with all three remainders. Now if $a=b$, then both $c$ and $d$ are different from $a$ (otherwise, we reduce the problem to the previous case). If $d\ne c$, $a$, $c$, $d$ are the three distinct remainders, and we have no possibility for $x$. So $c=d$.

a	a
	x
c	c

But this prevents the other remainder $m$ to appear in the middle row, leaving only two cells for three numbers, which is not possible. So, in both cases, one of the sums is a multiple of $3$.