Problem: Let $n > 1$ be an integer.  Find, with proof, all sequences
$x_1$, $x_2$, \dots, $x_{n-1}$ of positive integers
with the following three properties:
\begin{enumerate}[(a)]
 \ii $x_1 < x_2 < \dotsb < x_{n-1}$;
 \ii $x_i + x_{n-i} = 2n$ for all $i = 1, 2, \dots , n - 1$;
 \ii given any two indices $i$ and $j$ (not necessarily distinct)
 for which $x_i + x_j < 2n$,
 there is an index $k$ such that $x_i + x_j = x_k$.
\end{enumerate}

Solution: The answer is $x_k = 2k$ only, which obviously work,
so we prove they are the only ones.

Let $x_1 < x_2 < \dots < x_n$ be
any sequence satisfying the conditions.
Consider:
\[ x_1 + x_1 < x_1 + x_2 < x_1 + x_3 < \dots
  < x_1 + x_{n-2}. \]
All these are results of condition (c),
since $x_1 + x_{n-2} < x_1 + x_{n-1} = 2n$.
So each of these must be a member of the sequence.

However, there are $n-2$ of these terms,
and there are exactly $n-2$ terms greater
than $x_1$ in our sequence.
Therefore, we get the one-to-one correspondence below:
\begin{align*}
  x_2 &= x_1 + x_1 \\
  x_3 &= x_1 + x_2 \\
  & \vdotswithin= \\
  x_{n-1} &= x_1 + x_{n-2}
\end{align*}
It follows that $x_2 = 2x_1$,
so that $x_3 = 3x_1$ and so on.
Therefore, $x_m = m x_1$.
We now solve for $x_1$ in condition (b)
to find that $x_1=2$ is the only solution,
and the desired conclusion follows.