5. 

Answer: 10

Example: Let's break down 2015 into 11 numbers of 155 and one number of 310. In this case, the sum of the addends can range from $155 \times 1$ to $155 \times 10$.

We will prove that it is impossible to get fewer than 10 different numbers: first, note that 2015 is not divisible by 12, so we will have at least 2 different numbers. Arrange all the numbers in ascending order. First, select only the first number, then the first and second, and so on—finally, select the first 9 numbers—this gives us 9 different sums. Now, take the last 9 numbers—the new sum exceeds all previous sums, so there are at least 10 different numbers on the board.
5. The number 2015 was divided into 12 addends, after which all numbers that can be obtained by adding some of these addends (from one to nine) were written down. What is the minimum number of numbers that could have been written down?