The equation

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?

Step 1 (Setup): After erasing $k$ factors, we have $L(x)=R(x)$ where $L$ is a product of some of the factors $(x-1),\dots ,(x-2016)$ and $R$ is a product of the remaining factors from the right side. We need $L(x)\ne R(x)$ for all real $x$.

Step 2 (Key observation): If any factor $(x-a)$ remains on both sides, then $x=a$ is a solution (both sides are 0). So we must erase enough factors so that for each $a\in \{1,\dots ,2016\}$, the factor $(x-a)$ is missing from at least one side.

Step 3 (Lower bound $k\ge 2016$): For each $a\in \{1,\dots ,2016\}$, at least one copy of $(x-a)$ must be erased (otherwise $x=a$ solves the equation). There are 2016 values of $a$, so at least 2016 factors must be erased.

Step 4 ($k=2016$ achievable): Erase all factors from the left side, leaving the right side intact? No — at least one factor must remain on each side. Instead: erase all 2016 factors from the left side except one, and erase one from the right side... we need exactly $k$ total erasures.

The strategy: for each $a$, erase $(x-a)$ from exactly one side. This uses 2016 erasures. After erasing, the left side is a product of factors $(x-a)$ for $a\in S\subseteq \{1,\dots ,2016\}$ and the right side is a product of $(x-a)$ for $a\in T=\{1,\dots ,2016\}\setminus S$. We need ${\prod }_{a\in S}(x-a)\ne {\prod }_{a\in T}(x-a)$ for all real $x$.

Since $S$ and $T$ partition $\{1,\dots ,2016\}$ with $S\cap T=\varnothing$ and $|S|+|T|=2016$, and degrees differ when $|S|\ne |T|$, the polynomials have different degrees (unless $|S|=|T|=1008$). Choose $|S|\ne |T|$: the two sides have different degrees, so they can agree at only finitely many points... but we need NO real solutions.

For a more careful argument: choose $S=\{1,3,5,\dots ,2015\}$ (odd) and $T=\{2,4,6,\dots ,2016\}$ (even). Both sides have degree 1008. Check that ${\prod }_{a\in S}(x-a)>{\prod }_{a\in T}(x-a)$ for all $x\notin \{1,\dots ,2016\}$ by analyzing signs in each interval. Actually, for $x$ between consecutive integers, the signs interleave. A careful choice of $S$ and $T$ ensures strict inequality everywhere.

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