Generalisation: Given a board $M \times N$ and dominoes with sizes $1 \times a$ and $b \times 1$ ($M,N,a,b \in {\mathbf N}$ ). Prove that the board can be tiled with these dominoes if and only if $a|N$ or $b|M$.

Since $2, 3 \nmid 2003$ the answer will be no; I prove the generalization from which the initial problem follows. The trick behind this may be applied to many other problems (see here as well). Write $\epsilon^j \omega^k$ in the $(j, k)$th square of the board, where $\epsilon$ and $\omega$ are primitive roots of unity of orders $b, a$ respectively. Then, the sum of the numbers in the cells covered by any generalized domino is null (sums of roots of unity). Hence, the sum over the whole covered board is null, that sum being $\left( \sum_{j=1}^M \epsilon^j \right) \left( \sum_{k=1}^N \omega^k \right)$. Hence at least one of the two factors will be zero, meaning that $a | N$ or $b | M$. The converse is obvious.

Here's a solution not using complex numbers. Suppose $b\nmid M$ $\Rightarrow$ $M=pb + r$, $0<r<b$. Let's color $1$st row, $(b+1)$-th row, $(2b+1)$-th row, ... $(pb+1)$-th row. There are $(p+1)N$ colored squares. Let $X$ be the number of colored squares which are covered with horizontal rectangles, and $Y$ number of them covered with vertical rectangles. Also, $Y$ is the number of all vertical rectangles on the board (because each vertical rectangle can cover exactly one stripe). If a horizontal rectangle covers at least one colored square, then the horizontal rectangle lies on a stripe and covers $a$ squares $\Rightarrow$ $a|X$. Now, let's color $b$-th row, $2b$-th row, ... $pb$-th row. There are $pN$ colored squares. Again, number of squares covered with vertical rectangles must be $Y$. So, number of squares covered with horizontal rectangles must be $X-N$ (since there are $N$ less colored squares). We again can conclude that $a|X-N$. $a|X$, $a|X-N$ $\Rightarrow a|N$.

One using "my" technique (as described in [1]): Assume the board has been tiled with 1x2 and 1x3 tiles. Let us divide each square of the board (and tile) in four smaller, with side equal to 1/2 of the original. That gives us a 4006x4006 square and place the coordinate system in the center of the board, with axes parallel to the board sides. Centroids of original 1x2 tiles have odd ordinates and even abscissas, while centroids of 1x3 tiles have both abscissas and ordinates odd. Both sum of abscissas and sum of ordinates for all tiles' centroids has to be zero (the centroid of whole system - centroid of the board). Zero is even, so since the sum of abscissas has to be zero (even), number of 1x3 tiles has to be even. That would imply that 1x3 tiles cover an even number of squares, just like the 1x2 tiles, but the board has an odd number of squares - contradiction. One may see that this is in fact a weaker form of the complex numbers method - but I like it [1] H. Šiljak, Centroids and Tiling problems, Math. Reflections 5 (2009)

hsiljak wrote: Centroids of original 1x2 tiles have odd ordinates and even abscissas, while centroids of 1x3 tiles have both abscissas and ordinates odd. Maybe I'm misunderstanding, but, by my reckoning, odd and even are the other way round here. And where do you take into account that the 1x2 tiles weigh 2, and the 3X1 tiles weigh 3? Merlin

I'm getting rusty, it seems. It's true, I switched the parity by mistake - and now the parity argument won't work since it implies just that there is an even number of 2x1 tiles, right? You asked where is the mass - in the parity argument it didn't play a role, since odd (3) times odd (supposed coordinates of 3x1 tiles) is odd again. Now that the first argument crashed and burnt, we might use it, I guess - zero is divisible by 6 and so are the weighted coordinates of 3x1 tiles, hence the number of 2x1 tiles has to be divisible by 3. That would mean that 2003² is divisible by 3, but it isn't, hence the contradiction. Have I missed something again? It has unfortunately been a really long time since I stopped solving tiling problems, so no wonder I lost the edge.

@ hsiljak - i think that works now And thank you for introducing this new method

Color the columns red and blue, with the leftmost column being red. Then, consider $r-b$, which are the number of squares covered by tiles, and we claim that it's invariant mod 3. This is because a $1\times 2$ horizontal domino increases $r,b$ by 1, and a $3\times 1$ increases one by 3. Thus, \[0\equiv r-b \equiv 2003 \pmod{3}\]a contradiction, so you cannot tile.