Write on the board the equation

$$
(x-1)(x-2) \cdots(x-2016)=(x-1)(x-2) \cdots(x-2016)
$$

where each side has 2016 linear factors. What is the smallest positive value of $k$ for which it is true that we can omit exactly $k$ of the 4032 linear factors so that at least one linear factor remains on each side, and the resulting equation has no real roots?
Nagy Kartal's solution. As the first step, we show that $k>2015$. If we delete fewer than 2016 linear factors, there will be a linear factor that appears on both sides. Let this linear factor be $(x-i)$. Then $i$ will be a root of the equation, since in this case both sides would be 0.

Now we show that there is a solution for $k=2016$. Let this equation be:

$$
\begin{aligned}
& (x-1)(x-4)(x-5)(x-8)(x-9) \cdots(x-2013)(x-2016)= \\
= & (x-2)(x-3)(x-6)(x-7) \cdots(x-2011)(x-2014)(x-2015)
\end{aligned}
$$

Now let's consider the two sides as two functions. Let the left side be $g(x)$, and the right side $f(x)$. We will show that for all $x$, $f(x)>g(x)$. We examine this case by case.

1. If $x<g(x)$.

2. If $x>2016$, then similarly, it can be shown that $f(x)>g(x)$.
3. For $1 \leq x \leq 2016$.

a) If $x$ is an integer of the form $4m$ or $4m+1$, then $g(x)=0, f(x)>0$.

b) If $x$ is an integer of the form $4m+2$ or $4m+3$, then $g(x)<0, f(x)>0$.

c) If $x$ is not an integer and $4m<x<4m+1$, then $g(x)>0, f(x)>0$, but $f(x)>g(x)$.

d) If $4m+1<x<4m+2$, then $g(x)<0, f(x)<0$, but $|g(x)|<|f(x)|$, so $f(x)>g(x)$.

e) If $4m+2<x<4m+3$, then $g(x)<0, f(x)<0$, but $|g(x)|<|f(x)|$, so $f(x)>g(x)$.

f) If $4m+3<x<4m+4$, then $g(x)>0, f(x)>0$, but $|g(x)|<|f(x)|$, so $f(x)>g(x)$.

In this case, we can also break down the functions:

$$
\begin{array}{ll}
f_{1}(x)=(x-2)(x-3), & f_{2}(x)=(x-6)(x-7) \\
g_{1}(x)=(x-1)(x-4), & g_{2}(x)=(x-5)(x-8)
\end{array}
$$

$$
\cdots,
$$

It is easily verifiable that $f_{i}(x)>g_{i}(x)$. Thus, it is also true that $f(x)>g(x)$.

This shows that the right side is always greater than the left side, so the equation has no roots.

The solution: $k=2016$.