, 2022

Solution:

Let $n = 2023$. If $P(x) = x^{n} + a_{n-1} x^{n-1} + \cdots + a_{0}$, then let
$$R(x) = x^{n} P\left(1 - \frac{1}{x}\right) = (x-1)^{n} + a_{n-1}(x-1)^{n} x + \cdots + a_{0} x^{n}$$
Then, note that $Q(x) = P(x) - R(x)$ is a polynomial of degree at most $n$, and it has roots $1, 2, \ldots, n$, so we have $Q(x) = k(x-1) \cdots (x-n)$ for some real constant $k$. Now we determine $P(x)$ in terms of $Q(x)$. If $g(x) = 1 - 1 / x$, then $g(g(x)) = \frac{1}{1-x}$ and $g(g(g(x))) = x$. Therefore, we have
$$\begin{aligned} P(x) - x^{n} P\left(1 - \frac{1}{x}\right) & = Q(x) \\ P\left(1 - \frac{1}{x}\right) - \left(1 - \frac{1}{x}\right)^{n} P\left(\frac{1}{1-x}\right) & = Q\left(1 - \frac{1}{x}\right) \\ P\left(\frac{1}{1-x}\right) - \left(\frac{1}{1-x}\right)^{n} P(x) & = Q\left(\frac{1}{1-x}\right) . \end{aligned}$$
Adding the first equation, $x^{n}$ times the second, and $(x-1)^{n}$ times the third yields
$$2 P(x) = Q(x) + x^{n} Q\left(\frac{x-1}{x}\right) + (x-1)^{n} Q\left(\frac{1}{1-x}\right),$$
so
$$\begin{aligned} & P(x) = \frac{k}{2} \big((x-1)(x-2) \cdots (x-n) + (0 x - 1)(-1 x - 1) \cdots (-(n-1) x - 1) \\ & \qquad + (-1 x + 0)(-2 x + 1) \cdots (-n x + (n-1))\big) . \end{aligned}$$
Therefore,
$$P(-1) = \frac{k}{2} (-(n+1)! + 0 + (2n+1)!!)$$
Also, since $P$ is monic, we know that
$$1 = \frac{k}{2} (1 + 0 - n!)$$
so
$$P(-1) = \frac{(2n-1)!! - (n+1)!}{1 - n!}$$
Modulo $2027$, $(n+1)! = 2024! \equiv -1$ and $n! = 2023! \equiv 1$ (by Wilson's theorem and properties of factorials mod prime). Also, $(2n+1)!! \equiv 0$. So our answer is
$$\frac{1/2}{1 - 1/6} = \frac{3}{5} \equiv \frac{2030}{5} = 406$$