Problem 1159

First, note that up to a translation, we can assume that $x_{1}=0$.

Suppose first that the $x_{i}$ are integers. In this case, if $I$ is a subset of $\{1, \ldots, 2007\}$ of cardinality 7, then $11 \sum_{i \in I} x_{i} \equiv 0(\bmod 7)$ and thus, since 7 and 11 are coprime, $\sum_{i \in I} x_{i} \equiv 0(\bmod 7)$. We deduce that for all $i, x_{i} \equiv x_{0}=0$ $(\bmod 7)$, that is, all the $x_{i}$ are multiples of 7. Of course, the family of $\frac{x_{i}}{7}$ is still a solution to the problem. By infinite descent, we finally show that all the $x_{i}$ are zero, which is what we wanted.

Now let's handle the general case. We proceed by approximation. Take a family $\left(x_{i}\right)$ of reals. Let $N$ be a strictly positive integer and greater than all the inverses of the $\left|x_{i}\right|$ for $x_{i}$ non-zero. By applying the pigeonhole principle, we obtain a strictly positive integer $D$ and integers $p_{i}$ such that $\left|D x_{i}-p_{i}\right| \leqslant \frac{1}{155 N}$ for all $i$. Let's show that the family of the $p_{i}$ still satisfies the conditions of the statement. Take $I$ a subset of $\{1, \ldots, 2007\}$ of cardinality 7. Then we can find a subset $J$ of $\{1, \ldots, 2007\}$ of cardinality 11 such that:

$$11 \sum_{i \in I} D x_{i}=7 \sum_{j \in J} D x_{j}$$

Evaluate:

$$\begin{aligned} \left|11 \sum_{i \in I} p_{i}-7 \sum_{j \in J} p_{j}\right| & =\left|11 \sum_{i \in I}\left(p_{i}-D x_{i}\right)-7 \sum_{j \in J}\left(p_{j}-D x_{j}\right)\right| \\ & \leqslant 11 \sum_{i \in I}\left|p_{i}-D x_{i}\right|+7 \sum_{j \in J}\left|p_{j}-D x_{j}\right| \leqslant \frac{154}{155 N}<1 \end{aligned}$$

We deduce the expected equality, given that the number on the left-hand side is an integer. By the previous case, all the $p_{i}$ are zero and thus $\left|x_{i}\right| \leqslant\left|D x_{i}\right| \leqslant \frac{1}{155 N}$, which is impossible if $x_{i} \neq 0$ by the definition of $N$.