2. Let $x_{i} \in\{0,1\}(i=1,2, \cdots, n)$. If the function $f=f\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ takes values only 0 or 1, then $f$ is called an $n$-ary Boolean function, and we denote
$$
D_{n}(f)=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid f\left(x_{1}, x_{2}, \cdots, x_{n}\right)=0\right\} \text {. }
$$
(1) Find the number of $n$-ary Boolean functions;
(2) Let $g$ be an $n$-ary Boolean function, satisfying
$$
g\left(x_{1}, x_{2}, \cdots, x_{n}\right) \equiv 1+\sum_{i=1}^{n} \prod_{j=1}^{i} x_{j}(\bmod 2),
$$

Find the number of elements in the set $D_{n}(g)$, and find the largest positive integer $n$, such that
$$
\sum_{\left(x_{1}, x_{2}, \cdots, \cdots, x_{n}\right) \in D_{n}(g)}\left(x_{1}+x_{2}+\cdots+x_{n}\right) \leqslant 2017 .
$$
2. (1) Same as Question 1 (1) of Grade 1.
(2) Let $|D_{n}(g)|$ denote the number of elements in the set $D_{n}(g)$.
Obviously, $|D_{1}(g)|=1, |D_{2}(g)|=1$.
$$
\begin{array}{l}
\text { Also, } g\left(x_{1}, x_{2}, \cdots, x_{n}\right) \equiv 1+\sum_{i=1}^{n} \prod_{j=1}^{i} x_{j} \\
=\left(1+x_{1}\left(1+\sum_{i=2}^{n} \prod_{j=2}^{i} x_{j}\right)\right)(\bmod 2),
\end{array}
$$
$$
\begin{aligned}
& \text { Then }|D_{n}(g)|=2^{n-1}-|D_{n-1}(g)|(n=3,4, \cdots) . \\
& \text { Hence }|D_{n}(g)|=2^{n-1}-2^{n-2}+|D_{n-2}(g)|=\cdots \\
& =\sum_{k=0}^{n-1}(-1)^{k} \cdot 2^{n-1-k} \\
= & \frac{2^{n}+(-1)^{n+1}}{3} .
\end{aligned}
$$

Let $c_{n}=\sum_{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \in D_{n}(g)}\left(x_{1}+x_{2}+\cdots+x_{n}\right)$.
Notice that,
$$
\begin{array}{l}
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 0\right) \equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}(\bmod 2) \\
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 1\right) \\
\equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}+x_{1} x_{2} \cdots x_{n-1}(\bmod 2)
\end{array}
$$

If $x_{1} x_{2} \cdots x_{n-1}=1$, then
$$
\begin{array}{l}
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 1\right) \equiv n+1(\bmod 2) \\
=\left\{\begin{array}{l}
0, n \text { is odd; } \\
1, n \text { is even. }
\end{array}\right.
\end{array}
$$

If $x_{1} x_{2} \cdots x_{n-1}=0$, then
$$
\begin{array}{l}
g\left(x_{1}, x_{2}, \cdots, x_{n-1}, 1\right) \\
\equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}(\bmod 2)=0 \\
\Leftrightarrow g\left(x_{1}, x_{2}, \cdots, x_{n-1}\right) \\
\equiv 1+\sum_{i=1}^{n-1} \prod_{j=1}^{i} x_{j}(\bmod 2)=0
\end{array}
$$

Therefore, when $n \geqslant 2$,
$$
c_{n}=\left\{\begin{array}{l}
n+2 c_{n-1}+|D_{n-1}(g)|, n \text { is odd; } \\
2 c_{n-1}+|D_{n-1}(g)|-n, n \text { is even. }
\end{array}\right.
$$

Obviously, $c_{1}=1, c_{2}=1$.
When $n=2 m$,
$$
\begin{array}{l}
c_{2 m}=2 c_{2 m-1}+\frac{2^{2 m-1}+(-1)^{2 m}}{3}-2 m \\
=-\frac{1}{3}+\frac{4}{3} \times 2^{2 m-2}+(2 m-2)+4 c_{2 m-2} \\
=\frac{(3 m+1) 4^{m}-(6 m+1)}{9}
\end{array}
$$

Similarly, when $n=2 m+1$,
$$
\begin{array}{l}
c_{2 m+1}=2 m+1+2 c_{2 m}+\frac{2^{2 m}+(-1)^{2 m+1}}{3} \\
=\frac{1}{3}+\frac{4}{3} \times 2^{2 m-1}-(2 m-1)+4 c_{2 m-1} \\
=\frac{(6 m+5) 4^{m}+6 m+4}{9} .
\end{array}
$$

Therefore, from equations (2), (3), and
$$
\begin{array}{l}
c_{9}=828<3986=c_{11}, \\
c_{10}=1817<8643=c_{12},
\end{array}
$$

we know that the largest positive integer $n$ for which equation (1) holds is $n=10$.