Given an integer $n \geq 2$, let integers $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ satisfy: $0=a_{0}<a_{1}<\ldots<a_{n}=$ $2 n-1$. Find the minimum possible number of elements in the set $\left\{a_{i}+a_{j} \mid 0 \leq i \leq j \leq n\right\}$.
In summary, $S$ contains at least $3 n$ elements.
Note: The set $A+B=\{a+b \mid a \in A, b \in B\}$.
Fifty-seventh problem,
Solution: Let $S=\left\{a_{i}+a_{j} \mid 0 \leq \mathrm{i} \leq \mathrm{j} \leq \mathrm{n}\right\}$.
On one hand, take $a_{k}=k(0 \leq k \leq n-1)$, then $S=\{0,1,2, \ldots, 3 n-2\} \cup\{4 n-2\}$. At this point, $|S|=3 n$.
We now prove: $|S| \geq 3 n$ always holds.
First, $a_{0}<a_{1}<\ldots<a_{n}<a_{1}+a_{n}<a_{2}+a_{n}<\ldots<a_{n}+a_{n}$, these $2 n+1$ numbers all belong to $S$.
Let $A=\left\{a_{0}, a_{1}, a_{2}, \ldots, a_{n}\right\}, B=\{0,1, \ldots, 2 n-1\} \backslash A$, then $|B|=n-1$, and each element in $B \cup(B+$ $\{2 \mathrm{n}-1\})$ does not belong to $\left\{a_{0}, a_{1}, \ldots, a_{n}\right\} \cup\left\{a_{1}+a_{n}, a_{2}+a_{n}, \ldots, a_{n}+a_{n}\right\}$.
We now prove: For any $b \in B, b$ and $b+(2 n-1)$ must have one element belonging to $S$.
Assume $\mathrm{a}_{0}<\mathrm{a}_{1}<\ldots<\mathrm{a}_{\mathrm{t}}<\mathrm{b}<\mathrm{a}_{\mathrm{t}+1}<\ldots<\mathrm{a}_{\mathrm{n}}$.
If $\mathrm{b} \notin \mathrm{S}$, then for any $0 \leq \mathrm{j} \leq \mathrm{t}, \mathrm{b}-\mathrm{a}_{\mathrm{j}} \notin \mathrm{A}$, hence $\{0,1,2, \ldots, \mathrm{b}\}$ contains at least $\mathrm{t}+1$ elements not in $\mathrm{A}$.

If $b+(2 n-1) \notin S$, then for any $t+1 \leq j \leq n, b+(2 n-1)-a_{j} \notin A$, hence $\{b, b+$ $1, \mathrm{~b}+2, \ldots, 2 \mathrm{n}-1\}$ contains at least $\mathrm{n}-\mathrm{t}$ elements not in $\mathrm{A}$.

Thus, if $\mathrm{b}$ and $\mathrm{b}+(2 \mathrm{n}-1)$ are both not in $\mathrm{S}$, then $\{0,1, 2, \ldots, 2 \mathrm{n}-1\}$ contains at least $(\mathrm{t}+1)+$ $(n-t)-1=n$ elements not in $\mathrm{A}$, thus $|\mathrm{A}| \leq 2 \mathrm{n}-\mathrm{n}=\mathrm{n}$, which contradicts $|\mathrm{A}|=\mathrm{n}+1$.
Therefore, $|\mathrm{S}| \geq 2 \mathrm{n}+1+|\mathrm{B}|=3 \mathrm{n}$.