Problem: Find all integers $n \ge 2$ for which there exists
a sequence of $2n$ pairwise distinct points $(P_1, \dots, P_n, Q_1, \dots, Q_n)$
in the plane satisfying the following four conditions:
\begin{enumerate}[label={(\roman*)}]
  \ii no three of the $2n$ points are collinear;
  \ii $P_iP_{i+1} \ge 1$ for all $i=1,2,\dots,n$, where $P_{n+1} = P_1$;
  \ii $Q_iQ_{i+1} \ge 1$ for all $i=1,2,\dots,n$, where $Q_{n+1} = Q_1$; and
  \ii $P_i Q_j \leq 1$ for all $i=1,2,\dots,n$ and $j=1,2,\dots,n$.
\end{enumerate}

Solution: \paragraph{Answer.}
Even integers only.

\paragraph{Proof that even $n$ work.}
If we ignore the conditions that the points are pairwise distinct and form no
collinear triples, we may take
\[
  P_{2i+1}=(0.51, 0), \quad P_{2i}=(-0.51, 0), \quad
  Q_{2i+1}=(0, 0.51), \quad Q_{2i}=(0,-0.51).
\]
The distances $P_iP_{i+1}$ and $Q_iQ_{i+1}$ are $1.02 > 1$, while the distances
$P_iQ_j$ are $0.51\sqrt{2} < 1$. We may then perturb each point by a small
amount to ensure that the distance inequalities still hold and have the points
in general position.

\paragraph{Proof that odd $n$ do not work.}
The main claim is the following.

\begin{claim*}
  For $1\leq i\leq n$, points $Q_i$ and $Q_{i+1}$ must lie on opposite sides of
  line $P_1P_2$.
\end{claim*}

To isolate the geometry component of the problem,
we rewrite the claim in the following contrapositive form,
without referencing the points $Q_i$ and $Q_{i+1}$:
\begin{lemma*}
  Suppose $A$ and $B$ are two points such that $\max(P_1A, P_1B, P_2A, P_2B) \le 1$,
  and moreover $A$ and $B$ lie on the same side of line $P_1 P_2$.
  Further assume no three of $\{P_1, P_2, A, B\}$ are collinear.
  Then $AB < 1$.
\end{lemma*}

\begin{center}
\begin{asy}
  size(13cm);
  pair P_1 = (0,0);
  pair P_2 = (1,0);
  pair T = (0.5, 3^0.5/2);
  pair A = P_1 + dir(20);
  pair B = P_2 + 0.96*dir(145);
  pair X = extension(P_1, A, P_2, B);
  picture lp, rp;

  filldraw(lp, arc(P_1, P_2, T)--P_1--cycle, opacity(0.1)+yellow, blue+1.2);
  filldraw(lp, arc(P_2, T, P_1)--P_2--cycle, opacity(0.1)+yellow, blue+1.2);
  draw(lp, A--B, red);
  draw(lp, P_1--A, deepgreen);
  draw(lp, P_2--B, deepgreen);
  dot(lp, T);
  dot(lp, "$X$", X, dir(-90));
  dot(lp, "$P_1$", P_1, dir(-90));
  dot(lp, "$P_2$", P_2, dir(-90));
  dot(lp, "$A$", A, dir(A-P_1));
  dot(lp, "$B$", B, dir(245));

  pair A = P_1 + 0.9*dir(55);
  pair B = (0.85,0.03);
  filldraw(rp, arc(P_1, P_2, T)--P_1--cycle, opacity(0.1)+yellow, blue+1.2);
  filldraw(rp, arc(P_2, T, P_1)--P_2--cycle, opacity(0.1)+yellow, blue+1.2);
  draw(rp, A--B, red);
  draw(rp, P_1--A--P_2, deepgreen);
  draw(rp, P_1--B--P_2, dotted);
  dot(rp, "$P_1$", P_1, dir(-90));
  dot(rp, "$P_2$", P_2, dir(-90));
  dot(rp, "$A$", A, dir(260));
  dot(rp, "$B$", B, dir(65));

  add(lp);
  add(shift(1.2,0)*rp);
\end{asy}
\end{center}

\begin{proof}[Proof of lemma.]
  Suppose for the sake of contradiction
  that $A$ and $B$ lie on the same side of $P_1P_2$.
  The convex hull of these four points is either a quadrilateral or a triangle.
  \begin{itemize}
    \ii If the convex hull is a quadrilateral,
    assume WLOG that the vertices are $P_1P_2AB$ in order.
    Let $X$ denote the intersection of segments $P_1A$ and $P_2B$.
    Then \[ 1 + AB = P_1P_2 + AB < P_1X + XP_2 + AX + XB = P_1A + P_2B \leq 2. \]

    \ii Otherwise, assume WLOG that $B$ is in the interior of triangle $P_1P_2A$.
    Since $\angle P_1BA + \angle P_2BA = 360\dg-\angle P_1BP_2 > 180\dg$,
    at least one of $\angle P_1BA$ and $\angle P_2BA$ is obtuse.
    Assume WLOG the former angle is obtuse; then $AB < P_1A \leq 1$.
    \qedhere
  \end{itemize}
\end{proof}

\begin{remark*}
  Another proof of the lemma can be found by replacing segment $AB$ with
  the intersection of this line on the boundary of the blue region above,
  which does not decrease the distance.
  In other words, one can assume WLOG that $A$ and $B$ lie on either segment $AB$
  or one of the two circular arcs.
  One then proves that $AB \le 1$, and that for equality to occur,
  one of $A$ and $B$ must lie on segment $P_1 P_2$
  However, this approach seems to involve a fair bit more calculation.

  Yet another clever approach uses the trivia-fact that a
  \href{https://w.wiki/7VaD}{Reuleaux triangle} happens to have constant width.

  In any case, it's important to realize that this claim is \emph{not trivial};
  while it looks like it is easy to prove, it is not,
  owing to the two near-equality cases.
\end{remark*}

It follows from the claim that $Q_i$ is on the same side of line $P_1P_2$
as $Q_1$ if $i$ is odd, and on the opposite side if $i$ is even.
Since $Q_1 = Q_{n+1}$, this means the construction is not possible when $n$ is odd.

\begin{remark*}
  The fact that $n$ cannot be odd follows from Theorem 3 of
  \href{https://arxiv.org/pdf/1803.01822.pdf}{EPTAS for Max Clique on Disks and
  Unit Balls}. In the language of that paper, if $G$ is a unit ball graph, then
  the \emph{induced odd cycle parking number} of $\bar G$ is at most 1.

  In earlier versions of the proposed problem,
  the points were not necessarily distinct to make the even $n$ case nicer,
  but this resulted in annoying boundary conditions for the odd $n$ case.
\end{remark*}