We start with some basic observations. First note that the sequence $a_{1}, a_{2}, \ldots, a_{2023}$ is increasing and thus, since all elements are integers, $a_{n+1}-a_{n} \geqslant 1$. We also observe that $a_{1}=1$ and
$$a_{2}=\sqrt{\left(x_{1}+x_{2}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)}>2$$
by Cauchy-Schwarz inequality and using $x_{1} \neq x_{2}$. So, $a_{2} \geqslant 3$.

Now, we proceed to the main part of the argument. We observe that 3034 is about three halves of 2023. Motivated by this observation, we will prove the following.

Claim. If $a_{n+1}-a_{n}=1$, then $a_{n+2}-a_{n+1} \geqslant 2$.

In other words, the sequence has to increase by at least 2 at least half of the times. Assuming the claim is true, since $a_{1}=1$, we would be done since
$$\begin{aligned} a_{2023} & =\left(a_{2023}-a_{2022}\right)+\left(a_{2022}-a_{2021}\right)+\cdots+\left(a_{2}-a_{1}\right)+a_{1} \\ & \geqslant(2+1) \cdot 1011+1 \\ & =3034 \end{aligned}$$

We now prove the claim. We start by observing that
$$\begin{aligned} a_{n+1}^{2}= & \left(x_{1}+\cdots+x_{n+1}\right)\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n+1}}\right) \\ = & \left(x_{1}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n}}\right)+1 \\ & +\frac{1}{x_{n+1}}\left(x_{1}+\cdots+x_{n}\right)+x_{n+1}\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n}}\right) \\ \geqslant & a_{n}^{2}+1+2 \sqrt{\frac{1}{x_{n+1}}\left(x_{1}+\cdots+x_{n}\right) \cdot x_{n+1}\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n}}\right)} \\ = & a_{n}^{2}+1+2 a_{n} \\ = & \left(a_{n}+1\right)^{2}, \end{aligned}$$
where we used AM-GM to obtain the inequality. In particular, if $a_{n+1}=a_{n}+1$, then
$$\begin{equation*} \frac{1}{x_{n+1}}\left(x_{1}+\cdots+x_{n}\right)=x_{n+1}\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n}}\right) . \tag{1} \end{equation*}$$
Now, assume for the sake of contradiction that both $a_{n+1}=a_{n}+1$ and $a_{n+2}=a_{n+1}+1$ hold. In this case, (1) gives
$$\frac{1}{x_{n+2}}\left(x_{1}+\cdots+x_{n+1}\right)=x_{n+2}\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n+1}}\right) .$$
We can rewrite this relation as
$$\frac{x_{n+1}}{x_{n+2}}\left(\frac{1}{x_{n+1}}\left(x_{1}+\cdots+x_{n}\right)+1\right)=\frac{x_{n+2}}{x_{n+1}}\left(x_{n+1}\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n}}\right)+1\right) .$$
From (1) again, we conclude that $x_{n+1}=x_{n+2}$ which is a contradiction.

The trick is to compare $a_{n+2}$ and $a_{n}$. Observe that
$$\begin{aligned} a_{n+2}^{2}= & \left(x_{1}+\cdots+x_{n+2}\right)\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n+2}}\right) \\ = & \left(x_{1}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n}}\right)+\left(x_{n+1}+x_{n+2}\right)\left(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}\right) \\ & \quad+\left(x_{1}+\cdots+x_{n}\right)\left(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}\right)+\left(x_{n+1}+x_{n+2}\right)\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n}}\right) \\ \geqslant & a_{n}^{2}+\left(x_{n+1}+x_{n+2}\right)\left(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}\right) \\ & \quad+2 \sqrt{\left(x_{n+1}+x_{n+2}\right)\left(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}\right)\left(x_{1}+\cdots+x_{n}\right)\left(\frac{1}{x_{1}}+\cdots+\frac{1}{x_{n}}\right)} \\ = & a_{n}^{2}+\left(x_{n+1}+x_{n+2}\right)\left(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}\right)+2 a_{n} \sqrt{\left(x_{n+1}+x_{n+2}\right)\left(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}\right)}, \end{aligned}$$
where we used AM-GM to obtain the inequality. Furthermore, we have
$$\left(x_{n+1}+x_{n+2}\right)\left(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}}\right)>4$$
because $x_{n+1} \neq x_{n+2}$ by assumption. Therefore, it follows that
$$a_{n+2}^{2}>a_{n}^{2}+4+4 a_{n}=\left(a_{n}+2\right)^{2}$$
Because $a_{n+2}$ and $a_{n}$ are both positive integers, we conclude that
$$a_{n+2} \geqslant a_{n}+3$$
A simple induction gives $a_{2 k+1} \geqslant 3 k+a_{1}$ for every $k \geqslant 0$. Since $a_{1}=1$, it follows that $a_{2 k+1} \geqslant 3 k+1$. We get the desired conclusion for $k=1011$.