(5) Consider all non-empty subsets of the set $S=\{1,2, \cdots, 10\}$. If the number of even numbers in a non-empty subset is no less than the number of odd numbers, this subset is called a "good subset". The number of "good subsets" is ( ) .
(A) 631
(B) 633
(C) 635
(D) 637
Method 1: Let a "good subset" contain $i (i=1,2,3,4,5)$ even numbers, then the number of odd numbers can be $j=0,1, \cdots, i$. Therefore, the number of "good subsets" is
$$
\begin{aligned}
\sum_{i=1}^{5}\left(\mathrm{C}_{5}^{i} \sum_{j=0}^{i} \mathrm{C}_{5}^{j}\right)= & \mathrm{C}_{5}^{1}\left(\mathrm{C}_{5}^{0}+\mathrm{C}_{5}^{1}\right)+\mathrm{C}_{5}^{2}\left(\mathrm{C}_{5}^{0}+\mathrm{C}_{5}^{1}+\mathrm{C}_{5}^{2}\right) \\
& +\mathrm{C}_{5}^{3}\left(\mathrm{C}_{5}^{0}+\mathrm{C}_{5}^{1}+\mathrm{C}_{5}^{2}+\mathrm{C}_{5}^{3}\right)+\mathrm{C}_{5}^{4}\left(\mathrm{C}_{5}^{0}+\mathrm{C}_{5}^{1}+\mathrm{C}_{5}^{2}\right. \\
& \left.+\mathrm{C}_{5}^{3}+\mathrm{C}_{5}^{4}\right)+\mathrm{C}_{5}^{5}\left(\mathrm{C}_{5}^{0}+\mathrm{C}_{5}^{1}+\mathrm{C}_{5}^{2}+\mathrm{C}_{5}^{3}+\mathrm{C}_{5}^{4}+\mathrm{C}_{5}^{5}\right) \\
= & 637 .
\end{aligned}
$$

Therefore, the answer is D.
Method 2: The non-empty subsets of $S$ are $2^{10}-1=1023$ (sets). According to the number of even and odd numbers in the subsets, they can be divided into three categories: (1) even numbers outnumber odd numbers; (2) odd numbers outnumber even numbers; (3) the number of odd and even numbers is equal. Since the 10 elements in $S$ have an equal number of even and odd numbers, the number of subsets in (1) and (2) are equal. Now consider the third category, and consider the number of subsets containing $2, 4, 6, 8, 10$ elements, then the total number of subsets is
$$
\mathrm{C}_{5}^{1} \cdot \mathrm{C}_{5}^{1}+\mathrm{C}_{5}^{2} \cdot \mathrm{C}_{5}^{2}+\mathrm{C}_{5}^{3} \cdot \mathrm{C}_{5}^{3}+\mathrm{C}_{5}^{4} \cdot \mathrm{C}_{5}^{4}+1=251 .
$$

Therefore, the number of subsets in the first category is $\frac{1}{2}(1023-251)=386$.
Thus, the number of "good subsets" is $386+251=637$ (sets). Therefore, the answer is D.