Team Selection Tests

Because the polynomial $P(x)$ is non-constant and has the highest coefficient equal to $1$, there exists a constant $c$ such that $P(x)$ can take on all values above $[c, +\infty)$.
From equation (1), we get
$$f(f(x) + y + 2023f(y)) = x + 2024f(y), \quad \forall x \ge c, y \in \mathbb{R}. \quad (2)$$
We will prove $\lim_{x \to +\infty} |f(x)| = +\infty$. Indeed, assuming the existence of real numbers $M, m$ (we can assume $m \ge c$) such that $|f(x)| \le M$ for all $x \ge m$. Fix a real number $y_0$. For $x \ge m$, we have
$$-M + y_0 + 2023f(y_0) \le f(x) + y_0 + 2023f(y_0) \le M + y_0 + 2023f(y_0).$$
In addition, since $f$ is continuous on $\mathbb{R}$, there exists a real number $N$ such that $f(x) \le N$ for all real numbers $x$ belonging to the interval $[-M + y_0 + 2023f(y_0), M + y_0 + 2023f(y_0)]$. From here and from (2), we deduce
$$N \ge f(f(x) + y_0 + 2023f(y_0)) = x + 2024f(y_0),$$
or $N \ge x + 2024f(y_0)$ for all $x \ge m$, contradiction. Therefore $\lim_{x \to +\infty} |f(x)| = +\infty$.
From (2), it is easy to see that $f$ is injective on the domain $[c, +\infty)$. Since $f$ is a continuous function on $\mathbb{R}$, $f$ is a truly monotone function on the domain $[c, +\infty)$. Thus, there are two possible cases.

Case 1: $\lim_{x \to +\infty} f(x) = -\infty$. We will prove $\lim_{x \to -\infty} f(x) = +\infty$. Indeed, suppose there exist real numbers $L, \ell$ such that $f(x) \le L$ for all $x \le \ell$. Fix the real number $y_0$. Since $\lim_{x \to +\infty} f(x) = -\infty$, there exists a real number $x_0 \ge c$ such that $x_0 > L - 2024f(y_0)$ and $f(x_0) + y_0 + 2023 \le \ell$. Then, from (2), we have
$$L < x_0 + 2024f(y_0) = f(f(x_0) + y_0 + 2023f(y_0)) \le L.$$
From the received contradiction, we deduce $\lim_{x \to -\infty} f(x) = +\infty$. Because $\lim_{x \to +\infty} f(x) = -\infty$, $\lim_{x \to -\infty} f(x) = +\infty$ and $f$ is continuous on $\mathbb{R}$ so $f$ is an all-reflection on $\mathbb{R}$.
Now, fixing the real number $x_1 \ge c$. We see that there exists a real number $y_1$ such that $f(y_1) = -\frac{f(x_1)}{2023}$. Then, from equation (2), we deduce
$$x_1 - \frac{2024}{2023} f(x_1) = x_1 + 2024 f(y_1) = f(f(x_1) + y_1 + 2023f(y_1)) = f(y_1) = -\frac{f(x_1)}{2023},$$
therefore $f(x_1) = x_1$. Thus, $f(x) = x$ for all $x \ge c$, which is a contradiction because $\lim_{x \to +\infty} f(x) = -\infty$.

Case 2: $\lim_{x \to +\infty} f(x) = +\infty$. In this case, it is easy to see that $f$ is a strictly increasing function on the domain $[c, +\infty)$.
Now, suppose there are two real numbers $a, b$ such that $f(a) = f(b)$. Since $\lim_{x \to +\infty} f(x) = +\infty$ so there exists a real number $d \ge c$ such that $f(d) + a + 2023f(a) \ge c$ and $f(d) + b + 2023f(b) \ge c$. Then, from equation (2), we have
$$f(f(d) + a + 2023f(a)) = d + 2024f(a) = d + 2024f(b) = f(f(d) + b + 2023f(b)).$$
Because $f$ increases strictly over the domain $[c, +\infty)$, $f(d) + a + 2023f(a) = f(d) + b + 2023f(b)$, implies $a = b$. Thus, the function $f$ is one-to-one on $\mathbb{R}$. Because $f$ is continuous on $\mathbb{R}$ and $f$ is strictly increasing on the domain $[c, +\infty)$ so from here, we deduce that $f$ increases strictly on $\mathbb{R}$.
Next, we will prove $\lim_{x \to -\infty} f(x) = -\infty$. Indeed, suppose there exist real numbers $V, \nu$ such that $f(x) \ge V$ for all $x \le \nu$. Substituting $y = -f(x)$ into equation (2), we get
$$f(2023f(-f(x))) = x + 2024f(-f(x)), \quad \forall x \ge c. \quad (3)$$
Since $\lim_{x \to +\infty} f(x) = +\infty$, there exists a real number $x_2 \ge c$ such that $x_2 > f(2023f(\nu)) - 2024V$ and $f(x_2) \ge -\nu$. We have $-f(x_2) \le \nu$ so $f(-f(x_2)) \ge V$, follows output
$$x_2 + 2024f(-f(x_2)) \ge x_2 + 2024V.$$
There is again $-f(x_2) \le \nu$ so $f(-f(x_2)) \le f(\nu)$, infers
$$f(2023f(-f(x_2))) \le f(2023f(\nu)).$$
Combined with (3), we get
$$\begin{aligned} f(2023f(\nu)) &\ge f(2023f(-f(x_2))) = x_2 + 2024f(-f(x_2)) \\ &\ge x_2 + 2024L > f(2023f(\nu)). \end{aligned}$$
From the received contradiction, we deduce $\lim_{x \to -\infty} f(x) = -\infty$. We have $\lim_{x \to +\infty} f(x) = +\infty$, $\lim_{x \to -\infty} f(x) = -\infty$ and $f$ is continuous on $\mathbb{R}$ so $f$ is an all-reflection on $\mathbb{R}$. At here, doing the same substitution as case 1 above, we have $f(x) = x$ for all $x \ge c$. Now, in equation (2), fix the real number $x$ and choose $x \ge c$ enough so that $f(x) + y + 2023f(y) \ge c$. We have
$$\begin{aligned} x + y + 2023f(y) &= f(x) + y + 2023f(y) \\ &= f(f(x) + y + 2023f(y)) = x + 2024f(y). \end{aligned}$$
It follows that $f(y) = y$. Thus, we have $f(x) = x$ for all real numbers $x$. It is easy to verify that $f(x) = x$ indeed satisfies the condition of the problem.