$15 \cdot 11$ The number of sets of two or more consecutive positive integers whose sum is 100 is 
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(E) 5.
(24th American High School Mathematics Examination, 1973)
[Solution] In a set of consecutive integers, the number of elements can be odd or even. If it is odd, let it be $2n+1$, and their average $x$ is the central one. 
Then $(2n+1)x=100$, so $x=\frac{100}{2n+1}$. 
Thus $(2n+1)$ can only be 5 or 25.
If $2n+1=5$, then $x=20, n=2$, so these integers are
$$
18,19,20,21,22 \text{. }
$$

If $(2n+1)=25, x=4$ and $n=12$, this is impossible because the integers must be positive.

If the number of consecutive integers is even $2n$, let their average (between the middle pair of numbers) be represented by $x$, then $2nx=100$, i.e., $x=\frac{50}{n}$.
For $n=4$, i.e., $x=12\frac{1}{2}$, the middle pair of numbers are
$$
x-\frac{1}{2}=12, \quad x+\frac{1}{2}=13,
$$

so $9,10,11,12,13,14,15,16$ are these $2n=8$ integers.
For other integers $n$, there is no middle pair $\left(x-\frac{1}{2}\right)$ and $\left(x+\frac{1}{2}\right)$ that are positive integers.
In summary, the sets of positive integers whose sum is 100 are only the two mentioned above.
Therefore, the answer is $(B)$.