3. C.

Let the $k$ consecutive natural numbers be $n, n+1, \cdots, n+k-1$.
According to the problem,
$$
\begin{array}{l}
\frac{1}{2} k[n+(n+k-1)]=2000 \\
\Rightarrow(2 n+k-1) k=2^{5} \times 5^{3} .
\end{array}
$$

Since $2 n+k-1$ and $k$ should be one odd and one even, and
$$
\begin{aligned}
2 n+k & -1>k, \text { so, } \\
& (2 n+k-1, k) \\
& =(800,5),(160,25),(125,32) .
\end{aligned}
$$

Thus, there are 3 values for $k$.
3. Given a natural number $k>1$, the sum of $k$ consecutive natural numbers is 2000. Then the number of values of $k$ that satisfy the condition is ( ).
(A) 1
(B) 2
(C) 3
(D) 6