Does this set of numbers exist? For any $n \in \mathbb{N}$, is there any set of numbers $\{x_1, x_2, \ldots, x_n\} \subseteq \mathbb{R}$, such that:


*

*$i>j$ implies $x_i > x_j$,

*$i=j$ implies $x_i = x_j$,

*$i<j$ implies $x_i < x_j$,

*and $$\sum_{i=1}^n a_ix_i = \sum_{i=1}^n b_ix_i$$ only if, for any $i \in \{1,2,\ldots,n\}$, $a_i = b_i$, where $a_i$ and $b_i$ are indicator variables (either 0 or 1)?



My attempt
Perhaps something like this $\{2^1, 2^2, \ldots, 2^n\}$. Let's see for $n=10$, we get:
$$
\{2, 4, 8, 16, 32, 64, 128, 256, 512, 1024\}
$$
Now suppose that:


*

*I secretly chosen a subset $\mathcal{S} \subseteq \{2, 4, 8, 16, 32, 64, 128, 256, 512, 1024\}$, 

*and then I have computed the sum $\sum_{x \in \mathcal{X}}x$. 

*Finally, I have given you the sum (only the sum), which happened to be $16$.


Can you unambiguously identify the numbers that I have secretly chosen in $\mathcal{S}$? It seems that there is only one way to have $\sum_{x \in \mathcal{X}}x = 16$, which is if $\mathcal{S}=\{16\}$.
Now let's run a Python code and see if it can find any two subsets $\mathcal{S}_1, \mathcal{S}_2 \in \{2,4,\ldots,1024\}$, where the summation $\mathcal{S}_1=\mathcal{S}_2$, such that $\mathcal{S}_1 \ne \mathcal{S}_2$.
Code fails to find any contradiction. 
import random
import sys

def evaluation(rounds):
    # our candidate set that we hope to satisfy that requirement
    X = [2**i for i in range(1,11)]

    # let's test if there are any two differen subsets S1 and S2 tha their sums are
    # identical
    history = {}
    for r in range(0, rounds):
        # shuffle the master set to choose random from
        random.shuffle(X)

        # identify the last element ID of each set
        S_size = random.randrange(0, len(X)+1)

        # define the set
        S = set(X[0:S_size])

        # find sum
        S_sum = sum(S)

        # test 
        if S_sum not in history:
            history[S_sum] = S
        else: 
            if S == history[S_sum]:
                print('Satisfied:')
                print('   S1=%s' % sorted(list(S)))
                print('   S2=%s' % sorted(list(history[S_sum])))
                print('   sum=%s' % S_sum)
            else:
                print('NOT SATISFIED!')
                print('   S1=%s' % sorted(list(S)))
                print('   S2=%s' % sorted(list(history[S_sum])))
                print('   sum=%s' % S_sum)
                sys.exit(1)

evaluation(10000)

Yep, no contradiction found by code.
But: how to prove this?
 A: I think lots of such sets exist:
$x_i = 2^i$ or $x_i = 10.6^i$
Another interesting example:

Let $x_i$ be some non-consecutive elements from Fibonacci sequence. The uniqueness is by Zeckendorf's theorem.

I think it does not necessarily need $x_i$ to be increasing in order, as specified in the question. Because we are talking about finite sum here, order does not matter. But if you want to enforce increasing order, I see no issue, but just feel a bit unrelevant.
So the intuition is that if the sequence is sparse (not granular), and since the control you have is only $0$ or $1$; so you really do not have the luxury to miss out one item, while still have the sum equal. Also notice the question requires $\{x_i\}$ to be strictly increasing - no harm to enforce it, but not needed, because we are doing finite summation here, order of $\{x_i\}$ really does not matter.

EDIT Show that $x_i = 2^i$ is one sequence.
Proof. By induction. If $n=1$, easy to see it is true.
Suppose that it is also true for $n-1$. (*)
Now we want to show it is true for $n$.
The key here is to notice that Notice that $\sum_{i=1}^{n-1}x_i=2^{n}-2<x_{n}=2^n$.
i) if $a_n=b_n$, by (*) we need $a_i = b_i$ for all $i = 1,2,...,n-1$, in order for the summation to be equal.
ii) what if $a_n \ne b_n$? WOLG we let $a_n = 0,\,b_n = 1$, which results a difference of $2^n$ to the summation. We try to arrange the $a_i, b_i$ for $i=1,2,3,...,n-1$ to compensate the difference introduced by $a_n=0,b_n=1$. However, the maximum compensation we could get is $2^n-2$ by letting $a_1=a_2=...=a_{n-1} = 1,\, b_1=b_2=...=b_{n-1}=0$, but we still have a difference of $-2$ and could not make the summation equal.
Thus the only way to make the summation equal, is to have $a_i = b_i$ for all $i = 1,2,3,...,n$.
We are done. And one comment: it's worth noting that actually when $x_i=2^i$ is actually the binary-representation for an integer, and all those $a_i,b_i$, if we allow the summation to start with $0$, instead of $1$ in the question, they are exactly how you write the number in binary: ${a_1a_2a_3...a_n}_{(2)}$
One example: $19 = 10011_{(2)}$, and here $a_0=1,a_1=1, a_2=0, a_3=0, a_4=1$, and such representation is UNIQUE.
A: Let $x_i=B^{i-1}$ where $1<B\in \mathbb N.$ Suppose $(a_1,...,a_n)\ne (b_1,...,b_n)$ where each $a_i$ and $b_i$ is a non-negative integer not exceeding $B-1.$
Let $D= \sum_{i=1}^na_ix_i-\sum_{i=1}^nb_ix_i.$
Let $j$ be the largest $i$ such that $a_i\ne b_i.$ Then $D=\sum_{i=1}^j(a_i-b_i)x_i.$
If $j=1$ then  $|D|=|(a_1-b_1)x_1|=|a_1-b_1|=|a_j-b_j|\geq 1.$ 
If $j>1$ note that $|b_i-a_i|\leq B-1$ for every $i,$ and we have       $$ |D|= \left|(a_j-b_j)x_i- \sum_{i=1}^{j-1}(b_i-a_i)x_i\right|\geq$$ $$\geq |(a_i-b_j)x_j|-\left|\sum_{i=1}^{j-1}(b_i-a_i)x_i\right|\geq $$ $$\geq|a_j-b_j|x_j-\sum_{i=1}^{j-1}|b_i-a_i|x_i\geq $$ $$\geq  B^{j-1}-\sum_{i=1}^{j-1}(B-1)B^{i-1}=$$ $$=B^{j-1}-(B-1)\left(\frac {B^{j-1}-1}{B-1}\right)=1.$$
