4. On the board, there are 2010 non-zero numbers $\mathrm{a}_{1}$, $\mathrm{a}_{2}, \ldots \mathrm{a}_{2010}$ and the products of all pairs of adjacent numbers: $\mathrm{a}_{1} \cdot \mathrm{a}_{2}, \mathrm{a}_{2} \cdot \mathrm{a}_{3}, \ldots \mathrm{a}_{2009} \cdot \mathrm{a}_{2010}$. What is the maximum number of negative numbers that can be among the 4019 numbers?
4. Answer: 3014.

Solution. Consider the triples of numbers ( $\left.\mathrm{a}_{1}, \mathrm{a}_{2}, \mathrm{a}_{1} \cdot \mathrm{a}_{2}\right)$ ), ( $\left.\mathrm{a}_{3}, \mathrm{a}_{4}, \mathrm{a}_{3} \cdot \mathrm{a}_{4}\right), \ldots,\left(\mathrm{a}_{2009}, \mathrm{a}_{2010}, \mathrm{a}_{2009} \cdot \mathrm{a}_{2010}\right)$ ).

The product of the numbers in each triple is positive (as the product of squares of two non-zero numbers). Therefore, among the numbers in each triple, there is at least one positive number (the product of three negative numbers is negative). The specified triples of numbers do not overlap, so among the numbers written on the board, there are at least 1005 positive numbers. We will show that it is possible to choose such 2010 initial numbers that there will be 1005 positive numbers on the board. For this, we will make all the numbers with odd indices (a total of 1005) positive, and those with even indices negative. Then each product will be negative (as the product of two numbers with different signs), and the positive numbers will be only the 1005 initial numbers with odd indices. Therefore, the maximum number of negative numbers on the board will be 4019 $-1005=3014$.