China Team Selection Test

For a given integer $n \ge 2$, let $a_0, a_1, \ldots, a_n$ be integers satisfying $0 = a_0 < a_1 < \ldots < a_n = 2n-1$. We aim to find the smallest possible number of elements in the set $\{ a_i + a_j \mid 0 \le i \le j \le n \}$.

First, we prove that the set $\{ a_i + a_j \mid 1 \le i \le j \le n-1 \}$ takes all residues modulo $2n-1$. Consider the $2n$ numbers:
$$a_0 < a_1 < \cdots < a_{n-1} < a_n$$
and
$$r - a_0 > r - a_1 > \cdots > r - a_{n-1} > r - a_n$$
for any integer $0 \le r \le 2n-2$. By the Pigeonhole Principle, there must be two numbers that are congruent modulo $2n-1$. Since $a_i \not\equiv a_j \pmod{2n-1}$ for $1 \le i < j \le n-1$, there exist $1 \le i, j \le n-1$ such that $a_i \equiv r - a_j \pmod{2n-1}$, meaning $a_i + a_j \equiv r \pmod{2n-1}$.

Thus, the set $\{ a_i + a_j \mid 1 \le i \le j \le n-1 \}$ takes all residues modulo $2n-1$.

Returning to the original problem, we note that there are $2n+1$ distinct numbers:
$$a_0 + a_0 < a_0 + a_1 < \cdots < a_0 + a_{n-1} < a_n + a_0 < a_n + a_1 < \cdots < a_n + a_n,$$
which, modulo $2n-1$, take only $n$ different residues. Combining this with the fact that $\{ a_i + a_j \mid 1 \le i \le j \le n-1 \}$ takes all residues modulo $2n-1$, there are at least $n-1$ additional distinct numbers.

Therefore, the number of elements in the set $\{ a_i + a_j \mid 0 \le i \le j \le n \}$ is at least $2n + 1 + n - 1 = 3n$.

By setting $a_i = n-1$ for $1 \le i \le n-1$, we achieve $\left| \{ a_i + a_j \mid 0 \le i \le j \le n \} \right| = 3n$. Hence, the smallest possible number of elements in the set is $\boxed{3n}$.