4. Given a positive 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 $\left\{a_{i}+a_{j} \mid 0 \leqslant i \leqslant j \leqslant n\right\}$.
Hint: The minimum possible value is $3n$, start with an estimate.
First, prove: $a_{i}+a_{j}(0 \leqslant i \leqslant j \leqslant n-1)$ runs through a complete residue system modulo $2n-1$.

Since $a_{i}+a_{j}(0 \leqslant i \leqslant j \leqslant n)$ runs through a complete residue system modulo $2n-1$, thus, each of the other $n-1$ residue classes contains at least one number, resulting in $n-1$ numbers that are all distinct from the $2n$ numbers.

Additionally, $a_{n}+a_{n}=4n-2$ is the largest number and is necessarily distinct from the other numbers. Therefore, there are at least $2n+n-1+1=3n$ numbers. When taking $a_{i}=i(i=0,1, \cdots, n-1), a_{n}=2n-1$, the values of $a_{i}+a_{j}(0 \leqslant i \leqslant j)$ are
$$
0,1,2, \cdots, 2n-2,2n-1, \cdots, 3n-2,4n-2
$$
a total of $3n$ numbers.

In summary, the minimum possible value of the number of elements in the set $\left\{a_{i}+a_{j} \mid 0 \leqslant i \leqslant j \leqslant n\right\}$ is $3n$.