*1. Set $M$ consists of more than two consecutive natural numbers, and the sum of its elements is 1996. Such a set $M$ ( ).
(A) does not exist
(B) exists uniquely
(C) exists in two forms
(D) exists in more than three forms
$-、 1$. B.
Let $M=\{a, a+1, \cdots, a+n-1\}$. From $\frac{(a+a+n-1) n}{2}=1996$, we have $n(2 a+n-1)=8 \times 499$.

Since $n$ and $2 a+n-1$ have different parities, and $n<2 a+n-1$, and 499 is a prime number, we can only have $n=8$, then $2 a+n-1=2 a+7=499$, solving for $a$ gives $a=246$.