Looks like an IMO problem

Let $LHS = p(x)$ , $RHS = q(x)$ $LHS - RHS = r(x)$ . We have $\deg(p)=1009$ , and $\deg(q)=1008$ , so $\deg(r) = 1009$ . Therefore, we must have at most $1009$ roots for $r$ .

Notice that

$\begin{aligned} r(0)&=- 1 \times 3 \times 5 \times \ldots \times 2017 + 2 \times 4 \times 6 \times \ldots \times 2016 < 0\\ r(2018) &= 1 \times 3 \times 5 \times \ldots \times 2017 - 2 \times 4 \times 6 \times \ldots \times 2016 >0\\ \end{aligned}$

Furthermore, $r(2)>0, r(4)<0, r(6)>0, r(8)<0, \ldots, r(2016)<0$ . Thus, by the Intermediate Value Theorem, each of the 1009 disjoint intervals $(0,2), (2,4), (4,6), \ldots, (2016,2018)$ contains at least one root. We conclude that the number of roots of $r$ is $\boxed{1009}$ .

Let $f(x)=(x-1)(x-3)(x-5).....(x-2017) - (x-2)(x-4)(x-6).....(x-2016)$

From expansion there are at most $1009$ roots are there.

Let's start with $x=0$ .From simple observation

$f(0),f(1) <0 \,\, ; f(2),f(3) >0 \,\,; f(4) ,f(5) <0\,\,; f(6),f(7) >0\,\, \cdots \,\, f(2014),f(2015) >0\,\,; ; f(2016),f(2017) <0 \,\,; f(2018),f(2019) >0$

So, there exists roots between $(1,2) ; (3,4) ; .....(2015,2016);(2017,2018)$

Here number of ordered pairs are $1009$ .

The polynomial $f(x)$ has at most $1009$ roots so, we need not to check for negative values of $x$ .

Therefore, number of reals solutions is $1009$

Simple look at LHS there are 1009 terms so if you solve them highest power of x will be 1009 now look at that there it will be 1008 so no chance of cancellation. So 1009 sol. Possible as it's the degree and it will form a polynomial