Suppose that we form the random set $S \subseteq [n]$ by picking each person $i \in [n]$ to be in $S$ independently with probability $p > 0$ (to be chosen later). For each $i \in [n]$, let $X_i$ be the random variable counting the number of friends of $i$ that are in $S$. Note that $X_i \sim \mathrm{Binomial}(1000, p)$ for all $i$ (but also note that they are not independent!)
To apply the probabilistic method, we would like to show that probability of our desired event -- i.e. the random set $S$ is such that at least $n / 2023$ people in $S$ have exactly 2 friends in $S$ -- is positive, or with our notation, the following event holds with positive probability:
$$
\sum_{i \in [n]} 1_{i \in S, X_i = 2} \geq \frac{n}{2023} .
$$
This obviously looks quite complicated to compute exactly, but the key observation is that this event holds with positive probability if we can show a simpler relationship, namely that the expected number of people in $S$ with exactly two friends in $S$ is at least $n / 2023$. (In what follows, I'll assume that $n$ is divisible by $2023$ to simplify away the discreteness of the problem, but some slight modifications can be used for general $n$.)
By linearity of expectation, our aim is to show that
$$
\mathbb{E}\left[\sum_{i \in [n]} 1_{i \in S,\, X_i = 2}\right] = \sum_{i \in [n]} \mathbb{P}\left( i \in S, X_i = 2 \right) \geq \frac{n}{2023}
$$
Since the two events $i \in S$ and $X_i = 2$ are independent, $\mathbb{P}\left( i \in S, X_i = 2 \right) = \mathbb{P}\left( i \in S \right) \mathbb{P} \left(X_i = 2 \right)$. Hence, by symmetry, this is the same as showing that (for any $i$)
$$
f(p) := \mathbb{P}\left( i \in S \right) \mathbb{P} \left(X_i = 2 \right) \geq \frac{1}{2023} .
$$
Now perhaps you can try to verify that you can indeed choose $p \in (0, 1)$ such that this holds (hint: $p = 3 / 1001$ works) and wrap up the proof.
