Problem 4. Let $a_{1}, a_{2}, \ldots, a_{2005}, b_{1}, b_{2}, \ldots, b_{2005}$ be real numbers such that the inequality

$$
\left(a_{i} x-b_{i}\right)^{2} \geq \sum_{j=1, j \neq i}^{2005}\left(a_{j} x-b_{j}\right)
$$

holds true for every real number $x$ and all $i=1,2, \ldots, 2005$. Find the maximum possible number of the positive numbers amongst $a_{i}$ and $b_{i}, i=1,2, \ldots, 2005$.

Nazar Agakhanov, Nikolai Nikolov
4. We first prove that at least one of the numbers $a_{1}, a_{2}, \ldots, a_{2005}$ is not positive. To do this we assume the contrary and choose $i$ such that

$$
\frac{b_{i}}{a_{i}}=M=\max _{1 \leq j \leq 2005}\left(\frac{b_{j}}{a_{j}}\right)
$$

Then we can find $\varepsilon>0$ such that

$$
\left(a_{i} x-b_{i}\right)^{2}<\sum_{j=1, j \neq i}^{2005}\left(a_{j} x-b_{j}\right)
$$

for every $x \in(M, M+\varepsilon)$, a contradiction.

On the other hand, it is easy to see that if $a_{1}=a_{2}=\cdots=a_{2004}=$ $-a_{2005}=1$ and $b_{1}=b_{2}=\cdots=b_{2004}=b_{2005} \geq \frac{1001^{2}}{2}$ the given inequality is satisfied. Therefore the answer is 4009.