(i) Find the number of odd subsets and even subsets of $X$;
(ii) Find the sum of the sums of elements of all odd subsets of $X$.
Let $A$ be an odd subset of $X$. Consider the mapping $f$:
$$
\begin{array}{l}
A \longmapsto A-\{1\}, \text { if } 1 \in A ; \\
A \longmapsto A \cup\{1\}, \text { if } 1 \notin A .
\end{array}
$$

Clearly, $f$ is a mapping from odd subsets to even subsets. $f$ is injective, meaning that for different $A$, $f(A)$ is different.
$f$ is surjective, meaning that for every even subset $B$, there exists an $A$ such that $f(A)=B$. In fact, when $1 \in B$, let $A=B-\{1\}$; when $1 \notin B$, let $A=B \cup\{1\}$; then $f(A)=B$.

Thus, $f$ is a one-to-one correspondence from the family of odd subsets to the family of even subsets. Therefore, the number of odd subsets of $X$ is equal to the number of even subsets, both being $\frac{1}{2}|P(X)|=\frac{1}{2} \times 2^{n} = 2^{n-1}$.

As a corollary of (i), the number of odd subsets of $X$ that contain 1 is $2^{n-2}$ $\left(=\frac{1}{2} \times 2^{n-1}\right)$; the number of odd subsets that do not contain 1 is also $2^{n-2}$.
The sum of the sums of the elements of all subsets of $X$ is
$$
2^{n-1} \times(1+2+\cdots+n)=2^{n} n(n+1)
$$
(because each element $i$ appears in $2^{n-1}$ subsets).
Corresponding to the mapping $f$, each odd subset $A$ containing 1 has one more 1 than the even subset $B$, thus the sum of the elements is 1 more. Therefore, the sum of the sums of the elements of all odd subsets containing 1 ( $2^{n-2}$ subsets) is $2^{n-2}$ more than the sum of the sums of the elements of all even subsets not containing 1.

Similarly, the sum of the sums of the elements of all odd subsets not containing 1 is $2^{n-2}$ less than the sum of the sums of the elements of all even subsets containing 1.

Therefore, the sum of the sums of the elements of all odd subsets is equal to the sum of the sums of the elements of all even subsets, both being
$$
\frac{1}{2} \times 2^{n-2} n(n+1)=2^{(n-3)} n(n+1)
$$