Question 231, Positive integer $n \geq 2$, given pairwise distinct real numbers $a_{1}, a_{2}, \ldots, a_{n}$ forming set $S$, define $k(S)$ as the number of all distinct numbers of the form $a_{i}+2^{j}(i, j=1, 2, \ldots, n)$, for all sets $S$, find the minimum possible value of $k(S)$.

For all sets $S$, find the minimum possible value of $k(S)$.
For the set $S=\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$, define the set $T_{i}=\left\{a_{i}+2^{j} \mid j=1,2, \ldots, n\right\}$, where $i=1,2, \ldots, n$. Then
$k(S)=\left|T_{1} \cup T_{2} \cup \ldots \cup T_{n}\right|$
First, we prove that for any $i \neq j, \left|T_{i} \cup T_{j}\right| \leq 1$.
In fact, if $\left|T_{i} \cup T_{j}\right| \geq 2$, then there exist positive integers $p, q, r, s$ not greater than $n$, such that $a_{i}+2^{p}=a_{j}+2^{q}$, $a_{i}+2^{r}=a_{j}+2^{s}$. Without loss of generality, assume $a_{i}>a_{j}$, and $p<r$, then $p<q<s$, and $r<s$. Thus, we have $a_{i}-a_{j}=2^{q}-2^{p}=2^{s}-2^{r}$, rearranging gives $2^{q-p}-1=2^{r-p}\left(2^{s-r}-1\right)$. However, the left side of this equation is odd, while the right side is even, leading to a contradiction. Therefore, $\left|T_{i} \cup T_{j}\right| \leq 1$.
According to the principle of inclusion-exclusion, we have:
$$
k(S)=\left|T_{1} \cup T_{2} \cup \ldots \cup T_{n}\right| \geq \sum_{i=1}^{n}\left|T_{i}\right|-\sum_{1 \leq i<j \leq n}\left|T_{i} \cap T_{j}\right| \geq n \cdot n - C_{n}^{2} \cdot 1 = \frac{n(n+1)}{2}
$$

On the other hand, take $S=\left\{2^{1}, 2^{2}, \ldots, 2^{n}\right\}$, then the set $M$ of all numbers of the form $a_{i}+2^{j} (i, j=1,2, \ldots, n)$ is
$$
\begin{array}{l}
M=\left\{a_{i}+2^{j} \mid a_{i} \in S, j=1,2, \ldots, n\right\} \\
=\left\{2^{i}+2^{j} \mid i, j=1,2, \ldots, n\right\} \\
=\left\{2^{i}+2^{j} \mid 1 \leq i \leq j \leq n, i, j \in \mathbb{N}^{+}\right\}
\end{array}
$$

Clearly, $|M|=\frac{n(n+1)}{2}$, i.e., in this case, $k(S)=\frac{n(n+1)}{2}$.
In summary, the minimum possible value of $k(S)$ is $\frac{n(n+1)}{2}$.