Problem: Fix a real number $0 < t < \half$.
\begin{enumerate}
  \ii [(a)]
  Prove that there exists a positive integer $n$
  such that for every set $S$ of $n$ positive integers,
  the following holds:
  there exist distinct $x, y \in S$ and \emph{nonnegative} integer $m \ge 0$
  such that $|x-my| \le ty$.
  \ii [(b)]
  Determine whether there exists an infinite set $S$
  of positive integers such that the following holds:
  for any distinct $x, y \in S$ and \emph{positive} integer $m > 0$,
  we have $|x-my| > ty$.
\end{enumerate}

\end{enumerate}

Solution: \paragraph{Solution to (a).}
Assume not. Let $S = \left\{ s_1 < \dots < s_n \right\}$.
Consider
\[ 1 > \frac{s_1}{s_2} > \frac{s_1}{s_3} > \dots > \frac{s_1}{s_n} > t. \]
Note that two of the fractions above are within a factor of
$t^{1/(n-1)}$ of each other; taking $n$ large enough so that
$t^{1/(n-1)} \ge 1-t$ gives the conclusion.


\paragraph{Solution to (b).}
Yes, such a set $S$ exists.

To construct it, we employ a greedy algorithm.
Let $N$ be a large integer such that $t < 1/2-1/N$.
We define $S = \left\{ s_1 < s_2 < \dots \right\}$ inductively as follows.
\begin{itemize}
\ii First, let $s_1$ be any prime number exceeding $N$.

\ii Then, given $s_1$, \dots, $s_k$,
we let $s_{k+1}$ equal a prime number which is greater than $2s_k$,
and is congruent to $\tfrac{s_i-1}{2} \pmod{s_i}$ for $i = 1, \dots, k$.
This is possible by Chinese remainder theorem and Dirichlet theorem.
\end{itemize}

By construction, this works: if $i < j$ then $s_i / s_j < 1/2$
while $s_j / s_i$ has fractional part within $s_i^{-1}$ of $1/2$.