I've seen the following question in the brazilian olympiad for university students, and I couldn't solve it.
Thor and Loki play the game: Thor chooses an integer $n_1 \ge 1$ , Loki chooses $n_2 \gt n_1$, Thor chooses $n_3 \gt n_2$ and so on. Let $X$ be such that
$$X = \bigcup_{j\in\mathbb N^*} \left(\left[n_{2j-1},n_{2j}\right) \cap \mathbb Z \right)$$
and $$s= \sum_{n\in X} \frac {1}{n^2}$$
Thor wins if s is rational, and Loki wins if s is irrational. Determine who has got the winning strategy.
Loki should have the winning strategy. First note that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$, which is clearly irrational. Further, note that when Thor picks an integer he removes a set of rational numbers from the sum, thus lowering the highest possible value of the summation. Let $t_j=\sum_{n_{2j}\leq n\leq n_{2j+1}}\frac{1}{n^2}$ be the sum of the numbers removed at step $j$ (letting $n_0=1$). Note that $t_j$ is rational and that $s\leq \frac{\pi^2}{6} - \sum_{j=1}^m t_j$, where the right hand side is irrational.We will denote $s_j$ as the current sum at turn $j$ so that $s_j\to s$ as $j\to \infty$.
Loki can now impose some enumeration $q_1, q_2, ...$on $\mathbb Q$. On his $m^{th}$ turn, he picks the first rational in the list that is less than $\frac{\pi^2}{6} - \sum_{j=1}^m t_j$ and greater than $s_{m-1}$. Call this rational $q_i$. Note that if all remaining integers were picked after this turn, we would have $s=\frac{\pi^2}{6} - \sum_{j=1}^m t_j$, therefore Loki can pick some number of integers such that $q_i < s_m <\frac{\pi^2}{6} - \sum_{j=1}^m t_j$.
Continuing in this way, Loki can eliminate every possible rational number as being $s$. Thus $s$ will be irrational and Loki will win.
(Hopefully this makes sense. I'm not able to edit right now.)