Given an integer $n \geqslant 2$, let integers $a_{0}$, $a_{1}, \cdots, a_{n}$ satisfy
$$
0=a_{0}<a_{1}<\cdots<a_{n}=2 n-1 .
$$

Find the minimum possible number of elements in the set $\left\{a_{i}+a_{j} \mid 0 \leqslant i \leqslant j \leqslant n\right\}$.
Let $S=\left\{a_{i}+a_{j} \mid 0 \leqslant i \leqslant j \leqslant n\right\}$.
If we take $a_{k}=k(k=1,2, \cdots, n-1)$, then
$$
S=\{0,1, \cdots, 3 n-2,4 n-2\} .
$$

Therefore, $|S|=3 n$.
We now prove: $|S| \geqslant 3 n$ always holds.
It is easy to see that the following $2 n+1$ numbers
$$
a_{0}|X|$, then $Y \cap Z \neq \varnothing$.
Thus, there exists
$a_{j}=b_{i}+a_{n}-a_{k}$ or $a_{j}+a_{n}=b_{i}+a_{n}-a_{k}$.
Hence $b_{i}+a_{n}=a_{j}+a_{k} \in S$ or $b_{i}=a_{j}+a_{k} \in S$.
In conclusion, the minimum possible value of $|S|$ is $3 n$.