This problem was proposed by Burii.

rafaello wrote: Let $n$ be a positive integer and $t$ be a non-zero real number. Let $a_1, a_2, \ldots, a_{2n-1}$ be real numbers (not necessarily distinct). Prove that there exist distinct indices $i_1, i_2, \ldots, i_n$ such that, for all $1 \le k, l \le n$, we have $a_{i_k} - a_{i_l} \neq t$. W.l.o.g. $a_1\leq a_2\leq a_3\leq\cdots\leq a_{2n-1}$. Choose $i_j$ as the smallest index $i$ with $a_i-t\not\in\{a_{i_1},a_{i_2},\cdots,a_{i_{j-1}}\}$. Let $I=\{a_{i_1},a_{i_2},\cdots,a_{i_k}\}$ and $M=\{a_1, a_2, \cdots, a_{2n-1}\}\setminus I$. $f:M\to I, x\mapsto x-t$ is an injective mapping. Thus $|M|\leq |I|$ and $2|I|\geq 2n-1$ and $|I|\geq n$.

Draw a directed graph over the $2n-1$ indices so that $i \to j$ if and only if $a_i - a_j = t$. We want to show there exists a independent set of size $n$ in this graph. Note that this graph has no cycles since then we would have $t = 0$. In particular if we were to abandon the directedness of the graph, it would be a forest. Now assign each tree of the forest a appropriate bipartite coloring, and pick vertices whose color appear more times. Overall this ensures at least $\left \lceil \frac{2n-1}{2} \right \rceil = n$ vertices being selected. By the coloring hypothesis this is a independent set and thus we are done.

Very similar to above. Let $b_1 , b_2 , \dots , b_m$ denote the distinct numbers in $a_1 , a_2 , \dots , a_{2n-1}$. Construct a graph with vertices representing $b_1 , b_2 , \dots , b_m$ and draw an edge between $b_i$ and $b_j$ iff $|b_i - b_j| = t$. It is easy to see that this graph has no cycles, thus bipartite. Now since the graph is bipartite, one of it's parts have at least $\left \lceil \frac{2n-1}{2} \right \rceil = n$ elements, thus we are done.

Here is my solution: https://calimath.org/pdf/BalticWay2021-6.pdf And I uploaded the solution with motivation to: https://youtu.be/UYmvqRZFFzU

Create a graph, with $2n - 1$ vertices and $a_1, \dots, a_{2n - 1}$ assigned to these vertices. Draw an edge between any two vertices if the absolute value of their difference is $t$. The graph is bipartite, since it doesn't have an odd cycle (this is easy to prove), and by PHP, at least one of the two sets of vertices in the bipartite graph must be having at least $n$ vertices, so we're done.