Let the number of coefficients in the expansion of $(1+x)^{n}$ that leave a remainder of $r$ when divided by 3 be denoted as $T_{r}(n)$, $r \in\{0,1,2\}$. Calculate $T_{0}(2006), T_{1}(2006), T_{2}(2006)$.
Parse $(1+x)^{n}$ after expansion, there are a total of $n+1$ terms. Therefore,
$$
T_{0}(n)+T_{1}(n)+T_{2}(n)=n+1 .
$$

Expand $n$ in ternary:
$$
n=a_{0}+a_{1} \cdot 3+a_{2} \cdot 3^{2}+\cdots+a_{m} \cdot 3^{m},
$$

where $a_{i} \in\{0,1,2\}$.
Suppose (2) has $\alpha_{1}$ $a_{i}$'s that are 1, $\alpha_{2}$ $a_{i}$'s that are 2, and the rest are 0.
$$
\begin{aligned}
(1+x)^{3^{k}} & =\left(1+3 x+3 x^{2}+x^{3}\right)^{3^{k-1}} \equiv\left(1+x^{3}\right)^{3^{k-1}} \\
& \equiv \cdots \equiv\left(1+x^{3^{k-1}}\right)^{3} \equiv 1+x^{3^{k}}(\bmod 3) \\
(1+x)^{2 \cdot 3^{k}} & -\left((1+x)^{3^{k}}\right)^{2} \equiv\left(1+x^{3^{k}}\right)^{2}=1+2 x^{3^{k}}+x^{2 \cdot 3^{k}} \\
& \equiv 1-x^{3^{k}}+x^{2 \cdot 3^{k}}(\bmod 3) .
\end{aligned}
$$

Therefore,

where
$$
(1+x)^{n} \equiv \prod_{i=0}^{m}\left(1+g_{i}(x)\right)(\bmod 3) \text {, }
$$

when the degree of $f(x)$ is $<3^{i}$,
$$
f(x)\left(1+x^{3^{i}}\right)=f(x)+x^{3^{i}} f(x),
$$

the number of terms is twice the number of terms in $f(x)$, and the degree is $<3^{i+1}$.
$$
f(x)\left(1-x^{3^{i}}+x^{2 \cdot 3^{i}}\right)=f(x)-x^{3^{i}} f(x)+x^{2 \cdot 3^{i}} f(x),
$$

the number of terms is three times the number of terms in $f(x)$, and the degree is $<x^{3^{i+1}}$.
Therefore, starting from the lower degree, the number of terms in (3) is $2^{\alpha_{1}} \cdot 3^{\alpha_{2}}$, i.e.,
$$
T_{1}(n)+T_{2}(n)=2^{a_{1}} \cdot 3^{a_{2}} \text {. }
$$

If the coefficients of $f(x)$ are $\pm 1$, then in (4) or (5), the coefficients remain +1 or $-1$. The expansion of $\prod_{i=0}^{m}\left(1+g_{i}(x)\right)$ has $T_{1}(n)$ terms of +1 and $T_{2}(n)$ terms of -1.

Let $x=1$, then in the expansion of $\prod_{i=0}^{m}\left(1+g_{i}(x)\right)$, the $T_{1}(n)$ terms of +1 and the $T_{2}(n)$ terms of -1 cancel out to give $T_{1}(n)-T_{2}(n)$. And
$$
1+g_{i}(1)=\left\{\begin{array}{l}
2, \text { if } a_{i}=1, \\
1, \text { if } a_{i}=2,
\end{array}\right.
$$

Therefore,
$$
T_{1}(n)-T_{2}(n)=2^{a_{1}} .
$$

From (6), (7), and (1) we get
$$
\begin{array}{l}
T_{1}(n)=2^{\alpha_{1}-1}\left(3^{\alpha_{2}}+1\right), \\
T_{2}(n)=2^{\alpha_{1}-1}\left(3^{\alpha_{2}}-1\right), \\
T_{0}(n)=n+1-2^{\alpha_{1}} \cdot 3^{\alpha_{2}} .
\end{array}
$$

In particular, for $n=2006=(2202022)_{3}$, $\alpha_{1}=0, \alpha_{2}=5, T_{1}(n)=\frac{3^{5}+1}{2}=122$, $T_{2}(n)=\frac{3^{5}-1}{2}=121, T_{0}(n)=2007-3^{5}=1764$.