2.109 Let $n>2$ be a natural number. Try to construct a set consisting of $n$ distinct numbers $a_{1}, a_{2}, \cdots$, $a_{n}$, such that the set of sums of any two numbers
$$
S=\left\{a_{i}+a_{j} \mid 1 \leqslant i<j \leqslant n\right\}
$$

contains the fewest distinct numbers. Also, construct a set consisting of $n$ distinct numbers $b_{1}, b_{2}, \cdots$, $b_{n}$, such that the set of sums of any two numbers
$$
T=\left\{b_{i}+b_{j} \mid 1 \leqslant i<j \leqslant n\right\}
$$

contains the most distinct numbers.
[Solution] Let $0<a_{1}<a_{2}<\cdots<a_{n}$, then
$$
a_{1}+a_{2}<a_{1}+a_{3}<\cdots<a_{1}+a_{n}<a_{2}+a_{n}<\cdots<a_{n-1}+a_{n} \text {. }
$$

It can be seen that $S$ contains at least $2 n-3$ different numbers. Let
$$
a_{i}=i, i=1,2, \cdots, n,
$$

then $a_{i}+a_{j}=i+j, 3 \leqslant i+j \leqslant 2 n-1$. It is easy to see that, in this case,
$$
S=\{3,4, \cdots, 2 n-1\},
$$

which contains exactly $2 n-3$ distinct numbers.
Since the set $T$ contains at most $C_{n}^{2}$ distinct numbers, we can let
$$
b_{i}=10^{i-1}, i=1,2, \cdots, n .
$$

Then $b_{i}+b_{j}(i<j)$ is a $j$-digit number, where the $i$-th and $j$-th digits from the right are 1, and all other digits are 0. Clearly, for different pairs $(i, j)$, the resulting sum $b_{i}+b_{j}$ is also different. In this case, $T$ contains exactly $C_{n}^{2}$ distinct numbers.