Problem: The equation
\[ (x-1)(x-2)\dots(x-2016)=(x-1)(x-2)\dots (x-2016) \]
is written on the board, with $2016$ linear factors on each side.
What is the least possible value of $k$ for which it is possible to
erase exactly $k$ of these $4032$ linear factors so that at least
one factor remains on each side and the resulting equation
has no real solutions?

Solution: The answer is $2016$.
Obviously this is necessary in order to delete duplicated factors.
We now prove it suffices to deleted $2 \pmod 4$ and $3 \pmod 4$
guys from the left-hand side, and $0 \pmod 4$,
$1 \pmod 4$ from the right-hand side.

Consider the $1008$ inequalities
\begin{align*}
  (x-1)(x-4) &< (x-2)(x-3) \\
  (x-5)(x-8) &< (x-6)(x-7) \\
  (x-9)(x-12) &< (x-10)(x-11) \\
  &\vdots \\
  (x-2013)(x-2016) &< (x-2014)(x-2015).
\end{align*}
Notice that in all these inequalities, at most one of them
has non-positive numbers in it, and we never have both zero.
If there is exactly one negative term among the $1008 \cdot 2 = 2016$ sides,
it is on the left and we can multiply all together.
Thus the only case that remains is if $x \in (4m-2, 4m-1)$ for some $m$,
say the $m$th inequality.
In that case, the two sides of that inequality differ by a factor of at least $9$.

\begin{claim*}
  We have \[ \prod_{k \ge 0} \frac{(4k+2)(4k+3)}{(4k+1)(4k+4)} < e. \]
\end{claim*}
\begin{proof}
  [Proof of claim using logarithms]
  It's equivalent to prove
  \[ \sum_{k \ge 0} \log \left( 1 + \frac{2}{(4k+1)(4k+4)} \right) < 1. \]
  To this end, we use the deep fact that $\log(1+t) \le t$,
  and thus it follows from
  $\sum_{k \ge 0} \frac{1}{(4k+1)(4k+4)} < \half$,
  which one can obtain for example by noticing
  it's less than $\frac14\frac{\pi^2}{6}$.
\end{proof}
\begin{remark*}
  [Elementary proof of claim, given by Espen Slettnes]
  To avoid calculus as above, for each $N \ge 0$, note the partial product is bounded by
  \begin{align*}
    \prod_{k=0}^N \frac{(4k+2)(4k+3)}{(4k+1)(4k+4)}
    &=
    \frac 21 \cdot \left( \frac 34 \cdot \frac 65 \right)
      \cdot \left( \frac 78 \cdot \frac{10}{9} \right) \cdot \dots
      \cdot \frac{4N+3}{4N+4} \\
    &< 2 \cdot 1 \cdot 1 \cdot \dots \cdot \frac{4N+3}{4N+4}
    < 2 < e. \qedhere
  \end{align*}
\end{remark*}

This solves the problem, because then the factors being multiplied
on by the positive inequalities before the
$m$th one are both less than $e$, and $e^2 < 9$.
In symbols, for $4m-2 < x < 4m-1$ we should have
\[ \frac{(x-(4m-6))(x-(4m-5))}{(x-(4m-7))(x-(4m-4))}
  \times \dots \times \frac{(x-2)(x-3)}{(x-1)(x-4)} < e \]
and
\[ \frac{(x-(4m+2))(x-(4m+3))}{(x-(4m+1))(x-(4m+4))}
  \times \dots \times \frac{(x-2014)(x-2015)}{(x-2013)(x-2016)} < e \]
because the $(k+1)$st term of each left-hand side
is at most $\frac{(4k+2)(4k+3)}{(4k+1)(4k+4)}$, for $k \ge 0$.
As $e^2 < 9$, we're okay.