16 The sequences of sets $\left\{A_{n}\right\},\left\{B_{n}\right\}$ are defined as follows: $A_{1}=\varnothing, B_{1}=\{0\}, A_{n+1}$ $=\left\{x+1 \mid x \in B_{n}\right\}, B_{n+1}=\left(A_{n} \cup B_{n}\right) \backslash\left(A_{n} \cap B_{n}\right)$.
Find all positive integers $n$, such that $B_{n}=\{0\}$.
Define $f_{n}(x)=\sum_{b \in B_{n}} x^{b}, g_{n}(x)=\sum_{a \in A_{n}} x^{a}(n \geqslant 2)$.
Then we have $f_{1}(x)=1, g_{2}(x)=x, f_{2}(x)=1$.
Notice that the coefficients of each term in $f_{n}(x)$ and $g_{n}(x)$ are either 0 or 1. From the recursive definitions of $\left\{A_{n}\right\}$ and $\left\{B_{n}\right\}$, we have
$$
\begin{array}{l}
g_{n+1}(x)=x f_{n}(x), \\
f_{n+1}(x) \equiv f_{n}(x)+g_{n}(x)(\bmod 2),
\end{array}
$$

Thus, $f_{n+1}(x) \equiv f_{n}(x)+x f_{n-1}(x)(\bmod 2)$.
We will prove by induction that for any positive integer $m$, we have
$$
\begin{aligned}
f_{2 m}(x) & \equiv f_{m}\left(x^{2}\right)(\bmod 2) \\
f_{2 m+1}(x) & \equiv f_{m+1}\left(x^{2}\right)+x f_{m}\left(x^{2}\right)(\bmod 2)
\end{aligned}
$$

Proof: When $m=1$, $f_{2}(x) \equiv 1 \equiv f_{1}\left(x^{2}\right)(\bmod 2)$,
$$
\begin{aligned}
f_{3}(x) & \equiv f_{2}(x)+x f_{1}(x) \equiv 1+x \cdot 1 \\
& \equiv f_{2}\left(x^{2}\right)+x f_{1}\left(x^{2}\right)(\bmod 2) .
\end{aligned}
$$

Assume that the proposition holds for $m=k$, then for $m=k+1$, using the induction hypothesis and the recurrence relation (1), we have
$$
\begin{aligned}
f_{2 k+2}(x) & \equiv f_{2 k+1}(x)+x f_{2 k}(x) \\
& \equiv\left(f_{k+1}\left(x^{2}\right)+x f_{k}\left(x^{2}\right)\right)+x f_{k}\left(x^{2}\right) \\
& \equiv f_{k+1}\left(x^{2}\right)(\bmod 2) \\
f_{2 k+3}(x) & \equiv f_{2 k+2}(x)+x f_{2 k+1}(x) \\
& \equiv f_{k+1}\left(x^{2}\right)+x\left(f_{k+1}\left(x^{2}\right)+x f_{k}\left(x^{2}\right)\right) \\
& \equiv\left(f_{k+1}\left(x^{2}\right)+x^{2} f_{k}\left(x^{2}\right)\right)+x f_{k+1}\left(x^{2}\right) \\
& \equiv f_{k+2}\left(x^{2}\right)+x f_{k+1}\left(x^{2}\right)(\bmod 2)
\end{aligned}
$$

Thus, the proposition also holds for $m=k+1$.
By the principle of induction, the proposition holds for any positive integer $m$.
From (2) and (3), it is easy to see that the constant term of $f_{m}(x)$ is always 1, and the coefficient of the linear term of $f_{2 m+1}(x)$ is always 1.
When $n=2^{\alpha}(\alpha \geqslant 0)$, using (2) we have
$$
\begin{aligned}
f_{n}(x) & \equiv f_{2^{a}}(x) \equiv f_{2^{a-1}}\left(x^{2}\right) \\
& \equiv \cdots \equiv f_{1}\left(x^{2^{\alpha}}\right) \equiv 1(\bmod 2)
\end{aligned}
$$

When $n=2^{\alpha} \cdot(2 m+1)(m \geqslant 1, \alpha \geqslant 0)$, using (2) we have
$$
\begin{aligned}
f_{n}(x) & \equiv f_{2^{\alpha} \cdot(2 m+1)}(x) \\
& \equiv f_{2 m+1}\left(x^{2^{\alpha}}\right)(\bmod 2)
\end{aligned}
$$

Thus, the coefficient of the $x^{2^{\alpha}}$ term in $f_{n}(x)$ is 1, so $f_{n}(x) \neq 1(\bmod 2)$.
In summary, $B_{n}=\{0\}$ if and only if $m=2^{\alpha}(\alpha \geqslant 0, \alpha \in \mathbf{Z})$.