(13) (50 points) Let $k, n$ be given integers, $n > k \geqslant 2$. For any $n$-element set $P$, consider the sums of the elements of all $k$-element subsets of $P$, and let the set of these sums be $Q$. The number of elements in the set $Q$ is denoted as $C_{Q}$. Find the maximum value of $C_{Q}$.
(13) The maximum value of $C_{Q}$ is $\mathrm{C}_{n}^{k}$.
Since $P$ has $\mathrm{C}_{n}^{k}$ $k$-element subsets, it is obvious that $C_{Q} \leqslant \mathrm{C}_{n}^{k}$.
Below, we point out that for the set $P=\left\{2,2^{2}, \cdots, 2^{n}\right\}$, the corresponding $C_{Q}$ equals $\mathrm{C}_{n}^{k}$, i.e., the sum of elements of any two different $k$-element subsets of $P$ are not equal. Thus, the maximum value of $C_{Q}$ is $\mathrm{C}_{n}^{k}$.

In fact, if the set $P$ has two different $k$-element subsets $A=\left\{2^{r_{1}}\right.$, $\left.2^{r_{2}}, \cdots, 2^{r_{k}}\right\}, B=\left\{2^{s_{1}}, 2^{s_{2}}, \cdots, 2^{s_{k}}\right\}$, such that the sums of the elements of $A$ and $B$ are equal, then
$$
2^{r_{1}}+2^{r_{2}}+\cdots+2^{r_{k}}=2^{s_{1}}+2^{s_{2}}+\cdots+2^{s_{k}}=M .
$$

Since (1) can be viewed as the binary representation of the positive integer $M$, and since $r_{i}$ are distinct and $s_{i}$ are distinct, by the uniqueness of the binary representation of positive integers, we deduce from (1) that the set $\left\{r_{1}\right.$, $\left.r_{2}, \cdots, r_{k}\right\}$ must be the same as $\left\{s_{1}, s_{2}, \cdots, s_{k}\right\}$, thus the subsets $A=B$, which is a contradiction.
This proves our conclusion.