Existence of distribution for certain order statistics This is an open question: given a sequence of $n$ real numbers $x_1<x_2<\dots<x_n$, does there always exist a probability distribution, such that $\{x_i\}$ happens to be the $n$ expected order statistics of this distribution? 
In other words, can we always "reverse engineer" the distribution from its expected order statistics? Note that there is no restrictions on the distribution, i.e., it can be continuous, discrete, or whatever. I wonder if anything is known regarding this existence problem. 

(Edited 09/20/2017)
When sequence $\{x_i\}$ is unrestricted, the answer to above claim is no. This is shown by @Mateusz Kwaśnicki when $n=4$. However, under $n=4$, suppose $\{x_i\}$ satisfies the condition that $3(x_4−x_1)⩽7(x_3−x_2)$, then is there a method that can construct the distribution for which $\{x_1,x_2,x_3,x_4\}$ are the expected order statistics?
In other words, suppose $\{x_i\}$ satisfies the necessary conditions to be expected order statistics, is there a method to "reconstruct" the underlying distribution? Or is this too much to ask for?
 A: The answer is no when $n = 4$.
Let $X_1, X_2, X_3, X_4$ be a sample from a distribution with CDF $F$. Denote by $Y_1, Y_2, Y_3, Y_4$ the order statistics (that is, the non-decreasing rearrangement) of $X_1, X_2, X_3, X_4$, and let $x_j = \mathbb{E} X_j$. We will show that $3(x_4-x_1) \leqslant 7(x_3-x_2)$.
The CDFs $F_j$ of $Y_j$ are given by
$$\begin{aligned}
F_1 & = F^4 , & F_2 & = 4 F^3 - 3 F^4 , \\
F_3 & = 6 F^2 - 8 F^3 + 3 F^4 , \qquad & F_4 & = 4 F - 6 F^2 + 4 F^3 - F^4 .
\end{aligned}$$
Since $x_j = \mathbb{E} Y_j = \int_{-\infty}^\infty t \, dF_j(t)$, we have
$$
 3(x_4-x_1) - 7(x_3-x_2) = \int_{-\infty}^\infty t \, dG(t) ,
$$
where
$$
 G = 3F_4-7F_3+7F_2-3F_1 = 12 F - 60 F^2 + 96 F^3 - 48 F^4 = 12 F (1 - F) (1 - 2 F)^2.
$$
Observe that $t F(t)$ and $t (1 - F(t))$ converge to zero as $t \to -\infty$ and $t \to \infty$, respectively (because the distribution with CDF $F$ has finite mean). Therefore, $t G(t)$ converges to zero as $t \to \pm \infty$. Integration by parts leads to
$$
 3(x_4-x_1) - 7(x_3-x_2) = -\int_{-\infty}^\infty G(t) dt .
$$
However, $G = 12 F(1 - F)(1 - 2F)^2 \geqslant 0$, and so the right-hand side is non-positive, as claimed.

Interestingly, the answer to the original question is yes if $n = 3$, and it is enough to consider two-point distributions. Indeed, if $x_1 < x_2 < x_3$ and
$$\begin{aligned}
p & = \frac{x_3-x_2}{x_3-x_1}, & a & = \frac{3 x_1 x_3 - x_1^2 - x_1 x_2 - x_2^2}{3 (x_3 - x_2)} , & b & = \frac{x_2^2 + x_2 x_3 + x_3^2 - 3 x_1 x_3}{3 (x_2 - x_1)} ,
\end{aligned}$$
then $x_1, x_2, x_3$ are expected values of the order statistics of a sample of three random variables such that $\mathbb{P}(X = a) = p$ and $\mathbb{P}(X = b) = 1 - p$. This can be verified by a direct calculation, an important point is that $b - a = (x_3 - x_1)^3 / (3 (x_3 - x_2) (x_2 - x_1)) > 0$.
