Since there are 2016 common linear factors on both sides, we need to erase at least 2016 factors. We claim that the equation has no real roots if we erase all factors $(x-k)$ on the left-hand side with $k \equiv 2,3(\bmod 4)$, and all factors $(x-m)$ on the right-hand side with $m \equiv 0,1(\bmod 4)$. Therefore, it suffices to show that no real number $x$ satisfies
$$\prod_{j=0}^{503}(x-4 j-1)(x-4 j-4)=\prod_{j=0}^{503}(x-4 j-2)(x-4 j-3)$$
- Case 1. $x=1,2, \ldots, 2016$.
In this case, one side of (1) is zero while the other side is not. This shows $x$ cannot satisfy (1).
- Case 2. $4 k+1< x < 4 k+2$ for $0 \leqslant k \leqslant 503$ in this case.
So each term in the product lies strictly between 0 and 1, and the whole product must be less than 1, which is impossible.
- Case 3. $4 k+2 < x < 4 k+3$ for $0 \leqslant k \leqslant 503$ in this case.
So each term in the product lies strictly between 0 and 1, and the whole product must be less than 1, which is impossible.
- Case 4. $4 k+3 < x < 4 k+4$ for $0 \leqslant k \leqslant 503$ in this case.
So each term in the product lies strictly between 0 and 1, and the whole product must be less than 1, which is impossible.
- Case 5. $x < 1$ or $x > 2016$.
For $x < 1$, we have $|l(x)| > |r(x)|$. For $x > 2016$, we have $|l(x)| > |r(x)|$.
- Case 6. $x = 4k + 1$ or $x = 4k + 4$ for $0 \leq k \leq 503$.
For $x = 4k + 1$, we have $|l(x)| > |r(x)|$. For $x = 4k + 4$, we have $|l(x)| > |r(x)|$.
- Case 7. $x = 4k + 2$ or $x = 4k + 3$ for $0 \leq k \leq 503$.
For $x = 4k + 2$, we have $|l(x)| < |r(x)|$. For $x = 4k + 3$, we have $|l(x)| < |r(x)|$.

If $n \equiv 3(\bmod 4)$, one may leave $l(x)=(x-1)(x-2) \cdots\left(x-\frac{n+1}{2}\right)$ on the left-hand side and $r(x)=\left(x-\frac{n+3}{2}\right)\left(x-\frac{n+5}{2}\right) \cdots(x-n)$ on the right-hand side. For $x < 1$, we have $|l(x)| > |r(x)|$. For $1 < x < \frac{n+1}{2}$, we have $|l(x)| > |r(x)|$. For $x > \frac{n+1}{2}$, we have $|l(x)| > |r(x)|$.

If $n \equiv 1(\bmod 4)$, as the proposer mentioned, the situation is a bit more out of control. Since the construction for $n-1 \equiv 0(\bmod 4)$ works, the answer can be either $n$ or $n-1$. For $n=5$, we can leave the products $(x-1)(x-2)(x-3)(x-4)$ and $(x-5)$. For $n=9$, the only example that works is $l(x)=(x-1)(x-2)(x-9)$ and $r(x)=(x-3)(x-4) \cdots(x-8)$, while there seems to be no such partition for $n=13$.