Problem: A physicist encounters $2015$ atoms called usamons.
Each usamon either has one electron or zero electrons,
and the physicist can't tell the difference.
The physicist's only tool is a diode.
The physicist may connect the diode from any usamon $A$
to any other usamon $B$.  (This connection is directed.)
When she does so, if usamon $A$ has an electron and usamon $B$
does not, then the electron jumps from $A$ to $B$.
In any other case, nothing happens.  In addition,
the physicist cannot tell whether an electron jumps during any given step.
The physicist's goal is to isolate two usamons that she is 100\%
sure are currently in the same state.
Is there any series of diode usage that makes this possible?

Solution: The answer is no.
Call the usamons $U_1, \dots, U_m$ (here $m=2015$).
Consider models $M_k$ of the following form:
$U_1, \dots, U_k$ are all charged for some $0 \le k \le m$
and the other usamons are not charged.
Note that for any pair there's a model
where they are different states, by construction.

We can consider the physicist as acting
on these $m+1$ models simultaneously,
and trying to reach a state where there's a pair
in all models which are all the same charge.
(This is a necessary condition for a winning strategy to exist.)

But we claim that any diode operation $U_i \to U_j$
results in the $m+1$ models being an isomorphic copy
of the previous set.
If $i < j$ then the diode operation can be interpreted
as just swapping $U_i$ with $U_j$, which doesn't change anything.
Moreover if $i > j$ the operation never does anything.
The conclusion follows from this.

\begin{remark*}
  This problem is not a ``standard'' olympiad problem,
  so I can't say it's trivial.
  But the idea is pretty natural I think.

  You can motivate it as follows:
  there's a sequence of diode operations
  you can do which forces the situation to be one of the $M_k$ above:
  first, use the diode into $U_1$ for all other $U_i$'s,
  so that either no electrons exist at all or $U_1$ has an electron.
  Repeat with the other $U_i$.
  This leaves us at the situation described at the start of the problem.
  Then you could guess the answer was ``no'' just based on the fact
  that it's impossible for $n=2,3$
  and that there doesn't seem to be a reasonable strategy.

  In this way it's possible to give a pretty good description
  of what it's possible to do.
  One possible phrasing:
  ``the physicist can arrange the usamons in a line such that
  all the charged usamons are to the left of the un-charged usamons,
  but can't determine the \emph{number} of charged usamons''.
\end{remark*}