Is it the following set: $ \{1000,1,2,4,8,16,32,64,128,256,512\}$ and $ n=11$ ?

mszew wrote: Is it the following set: $ \{1000,1,2,4,8,16,32,64,128,256,512\}$ and $ n = 11$ ? There are many variants. For example minimal summ had $ 1,2,4,8,16,32,63,125,250,500,999.$

Of course those sets work. Now a proof that 10 doesn't work.

Is it correct? Suppose that $ x_1 < x_2 < \cdots < x_9 < x_{10}$ works, there are $ 2^{10} - 1 = 1023$ possible sums. If $ x_7 < 500$ then there are more than $ 22$ useless sums because all the sums of one or two different elements of $ \{x_1,\cdots,x_7\}$ are less than $ 1000$ $ (C_7^2 + C_7^1 = 28)$ then $ x_7 \ge 501$ But $ x_7 + x_8 + x_9 + x_{10} > 2000$ is a useless sum and all the sums with these four numbers, which are $ 2^6 = 64 > 22$ contradiction

I didn't check the numbers, but the reasoning works for sure.

mszew wrote: But $ x_7 + x_8 + x_9 + x_{10} > 2000$ is a useless sum and all the sums with these four numbers, which are $ 2^6 = 64 > 22$ contradiction there is a mistake: if $ x_1 <0$, then $ x_1+x_7 + x_8 + x_9 + x_{10}$ may less than 2000 and become a useful sum. my solution: since there are at most 22 useless sums, if there exist $ x_i$ not less than 1000 , then for any sum S among the other nine numbers, at least one of $ S$ and $ S+x_{i}$ is an useless sum. so there are at least 512 useless sums, a contradiction. if there exist i<>j such that $ x_{i}=x_j$, then for any sum S among the other eight numbers, one of $ S+x_{i}$ and $ S+x_{j}$ is an useless sum. so there are at least 256 useless sums, a contradiction, too. so for all i, $ x_{i}<1000$. thus the 10 sums of one number are useless. then there are at most 12 useless sums of at least two numbers if there are at least six numbers <500, then there are \[ {6 \choose 2}\] =15 useless sums of two numbers, a contradiction. if there are at least six numbers >=500, since there are at most one 500, so the sum of any four numbers among the six numbers is more than 2000. so there are \[ {6 \choose 4}\] =15 useless sums of four numbers, a contradiction. the rest case is that there are exactly 5 numbers <500 and 5 numbers >=500. in this case, for the "<500" numbers, there are \[ {5 \choose 2}\] =10 useless sums of any two numbers among them; for the ">=500" numbers, \[ {5 \choose 4}\] =5 useless sums of any four numbers among them. and 10+5>12, a contradiction.

$2^0 , 2^1 , ... ,2^{10}$ satisfies the problem