3. The number of positive integers $n$ such that $2 n+1$ divides $n^{3}+2$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
3. (C).

Since $2 n+1$ is coprime with 2, the problem is equivalent to $\mathrm{2} \frac{2^{3}\left(n^{3}+2\right)}{2 n+1}$ $=\frac{(2 n)^{3}+16}{2 n+1}=\frac{(2 n+1-1)^{3}+16}{2 n+1}$ being an integer, which is also equivalent to $\frac{16-1}{2 n+1}=\frac{3 \times 5}{2 n+1}$ being an integer. Therefore, $2 n+1=3,5,15$, i.e., $n=1$, 2, 7