3. The number of positive integers $n$ that satisfy $\mathrm{C}_{n}^{i}$ is coprime with 2007 for all $i=0,1, \cdots, n$ is ( ) .
(A) 3
(B) 5
(C) 7
(D) 9
3. D.

If the prime $p$ is coprime with all $\mathrm{C}_{n}^{i}(i=0,1, \cdots, n)$, since the maximum power of the prime factor $p$ in $\mathrm{C}_{n}^{i}$ is
$$
\sum_{k=1}^{+\infty}\left(\left[\frac{n}{p^{k}}\right]-\left[\frac{i}{p^{k}}\right]-\left[\frac{n-i}{p^{k}}\right]\right),
$$

then for every $i$ we have
$$
\left[\frac{n}{p^{k}}\right]=\left[\frac{i}{p^{k}}\right]+\left[\frac{n-i}{p^{k}}\right],
$$

which means that $i+(n-i)=n$ does not cause a carry in the $p$-ary representation. Thus, in the $p$-ary representation of $n$, all digits except the leading digit are $p-1$.
$$
\begin{array}{l}
\text { Let } f(r, p, m) \\
=r p^{m}+(p-1) p^{m-1}+\cdots+(p-1) p+(p-1) \\
=(r+1) p^{m}-1 .
\end{array}
$$

Then $n=1,2, \cdots, p-1, f(r, p, m)$, where $r=0,1, \cdots, p-2, m$ is any positive integer.

If a positive integer with exactly two distinct prime factors $p$ and $q (p<q)$ is coprime with all $\mathrm{C}_{n}^{i}(i=0,1, \cdots, n)$, then $n$ is the common positive integer in $1,2, \cdots, p-1, f\left(r_{1}, p, m_{1}\right)$ and $1,2, \cdots, q-1, f\left(r_{2}, p, m_{2}\right)$, where $r_{1}=0,1, \cdots, p-2, r_{2}=0,1, \cdots, q-2, m_{1}, m_{2}$ are any positive integers.
Since $\left(r_{1}+1\right) p^{m_{1}}-1 \neq\left(r_{2}+1\right) q^{m_{2}}-1$, then
$$
n=1,2, \cdots, p-1 \text { and }\left(r_{1}+1\right) p^{m_{1}}-1 \text {, }
$$

where $\left(r_{1}+1\right) p^{m_{1}}-1$ satisfies
$$
\left(r_{1}+1\right) p^{m_{1}}-1 \leqslant q-1 \text {. }
$$

Since $2007=3^{2} \times 223,\left(r_{1}+1\right) 3^{m_{1}}-1 \leqslant 223-1$, so $n=1,2,5,8,17,26,53,80,161$.
Therefore, the number of $n$ that satisfy the condition is 9.