A possible way to provide each vase with a note is to put in vase $1$ a note with $1$, in vase $2$ a note with $2$, in vase $3$ a note with $3$, $\ldots$, and in vase $2023$ a note with $2023$. We will use induction to show that this is the only distribution. Note that for a valid distribution it does not matter in which order we fill the vases.

Take a look at vase $1$. Suppose we put a note in it with $a$. Then in vase $a$ we put a note with $b$, such that $(a + b)/2 = 1$ or $a + b = 2$. Since $a$ and $b$ are positive integers, it must hold that $a = b = 1$. So the two vases here were the same and vase $1$ contains a note with $1$.

For the induction step, we assume that the first $n-1$ vases each contain a note with the number of the vase. We want to show that in vase $n$ we have to put a note with $n$. Suppose in vase $n$ we put a note with $a$. If $a < n$, then vase $a$ also contains a note with $a$ because of the induction hypothesis. It must then hold that $(a+a)/2 = n$, but this contradicts $a < n$. We conclude that $a \ge n$.
If $a > n$, we do not know yet which number has to go on the note in vase $a$. Call this number $b$. Then it must hold that $(a+b)/2 = n$, and so $b < n$. But then we find a contradiction if we were to consider vase $a$ first: in it we find a note with $b$, and in vase $b$ we then find, because of the induction hypothesis, another note with $b$. However, $(b+b)/2 \ne a$, because $a > n$ and $b < n$.
We conclude that $a = n$ must hold: in vase $n$ we also put a note with $n$ on it. Induction now gives that in each vase we put a note with the number of the vase on it. So this is the only possible distribution of the notes.