9.6. Answer. A maximum of 35 different numbers.

The sum of a blue and a red number can be a natural number from $1+2=3$ to $19+20=39$ inclusive, so there cannot be more than 37 different numbers. Moreover, note that one of the numbers $3,4, . ., 13$ must not be the sum of a blue and a red number. Otherwise, if the number 1 is blue, then 3 can only be represented as $1+2$, so 2 is red. The number 4 can only be represented as $1+3$, so 3 is also red. Next, $5=1+4=2+3$, but 2 and 3 are red, so only the first representation is valid, hence 4 is also red, and so on, we get that all 11 numbers $2,3, . ., 12$ are red, which is impossible. Similarly, we show that if all the numbers $29,30, \ldots, 39$ are sums of a blue and a red number, then 11 numbers $9, . ., 19$ are of the same color. Thus, there cannot be more than $37-2=35$ different numbers. One possible example where there are exactly 35 is as follows: $1,11,12, \ldots, 19$ are blue, and $2,3, \ldots, 10,20$ are red.

Grading remarks. Only the correct answer - 0 points. The correct answer and an example where it is achieved - 3 points. A correct proof of the maximum number 35 is worth 4 points. Attempts to solve where other estimates, different from 35, are proven, are graded 0 points.
6. Among all natural numbers from 1 to 20 inclusive, some 10 numbers were painted blue, and the other 10 - red, then all possible sums of pairs of numbers, one of which is blue and the other is red, were counted. What is the maximum number of different sums that can be among the hundred obtained numbers?