5. Among the 100 integers $\{1,2, \cdots, 100\}$, if $k$ numbers are chosen, such that among these $k$ numbers, there are always two numbers whose sum equals the sum of two other different numbers. Then, the smallest value of $k$ that satisfies this condition is ( ).
(A) 21
(B) 24
(C) 27
(D) 30
5. (A).

Among the first 100 natural numbers, the smallest sum of any two different numbers is 3, and the largest is $19 \%$, making 197 different values.

If we take $k$ numbers, each number can form a different sum with the other $k-1$ numbers. $k$ numbers can form $\frac{1}{2} k(k-1)$ sums (these sums are not necessarily distinct. Since the sums $a+b$ and $b+a$ are the same, they should be divided by 2). By the pigeonhole principle, when $k$ satisfies $\frac{1}{2} k(k-1) > 197$, at least two different pairs of numbers will have the same sum: $a+b = c+d$, and $a, b, c, d$ will not have any repeated numbers (obviously $a \neq b$. If $a=c$, then $b=d$, which would not be two different pairs). Taking $k=21$, the left side of the inequality is 210, so the inequality holds.