Let $N = 2023$. We assume that the equation $x^3 - Nx^2 - Nx + r = 0$ has three rational solutions $\frac{a}{k}$, $\frac{b}{k}$, $\frac{c}{k}$, where $a, b, c$ are integers and $k$ is a positive integer with $\gcd(a, b, c, k) = 1$. According to Vieta we have $\frac{a}{k} + \frac{b}{k} + \frac{c}{k} = N$ and $\frac{b}{k} \cdot \frac{c}{k} + \frac{a}{k} \cdot \frac{c}{k} + \frac{a}{k} \cdot \frac{b}{k} = -N$. This is equivalent to
$$\begin{aligned} a+b+c &= kN & \Rightarrow \quad a^2+b^2+c^2+2(bc+ac+ab) &= k^2N^2 \\ bc+ac+ab &= -k^2N & \Rightarrow \quad a^2+b^2+c^2 &= k^2N^2+2k^2N = k^2N(N+2). \end{aligned}$$

In a next step, we recognize that $k$ cannot be even. If it were, we would have $a^2+b^2+c^2 \equiv 0 \pmod 4$, from which we obtain that $a, b, c$ are all even, as 0 and 1 are the only quadratic residues modulo 4. This contradicts the assumption that $\gcd(a, b, c, k) = 1$.
For odd values of $k$, we have $k^2 \equiv 1 \pmod 8$. Furthermore, we have $N = 2023 \equiv 7 \pmod 8$. From this, we obtain $k^2 N(N+2) \equiv 1 \cdot 7 \cdot 1 \equiv 7 \pmod 8$. The sum of three perfect squares can never be congruent to 7 modulo 8, which can easily be verified by adding all possible combinations (the only quadratic residues modulo 8 are 0, 1 and 4). It follows that the above equation can never have three rational solutions.

(Josef Greilhuber) $\square$