Greece — Selection Examination

Let he choose at the beginning $\alpha_1$ from the seven pieces and each of them into seven pieces. Then he will have totally $7 - \alpha_1 + 7\alpha_1 = 7 + 6\alpha_1$ pieces of paper. Suppose that in the next step he chooses $\alpha_2$ pieces of paper and cuts each of them into seven pieces. Then he will have totally $7 + 6\alpha_1 - \alpha_2 + 7\alpha_2 = 7 + 6(\alpha_1 + \alpha_2)$ pieces of paper. If he continues this procedure $\kappa$ times, then he will have totally $7 + 6(\alpha_1 + \alpha_2 + \dots + \alpha_\kappa)$ pieces of paper. Therefore we are looking for the value of $\kappa$ satisfying the equation
$$7 + 6(\alpha_1 + \alpha_2 + \dots + \alpha_\kappa) = 2009 \Rightarrow 6(\alpha_1 + \alpha_2 + \dots + \alpha_\kappa) = 2002,$$
which is absurd, because $2002$ is not divided by $6$. Hence it is not possible for him to have some time $2009$ pieces of paper.