16.2.19* Between two 1s, there are 1991 nines, i.e.,
$$
1 \underbrace{99 \cdots 991}_{1001 \uparrow 9} \text {, }
$$

Try to find:
(1) Where should the “+” sign be placed between which two digits to make the sum of the two numbers the smallest?
(2) Where should the “ $\times$ ” sign be placed between which two digits to make the product of the two numbers the largest?
Parse: In (1), the number containing the first $m$ nines equals $2 \cdot 10^{m}-1$, and the remaining digits form the number $10^{1992-m}-9$.
(1) and $S(m)=2 \cdot 10^{m}+10^{1992-m}-10$ satisfy the estimate: $S(996) \leqslant 2 \cdot 10^{996}+$ $10^{996}=3 \cdot 10^{996}$. When $m \geqslant 997$, $S(m)>10^{m} \geqslant 10 \cdot 10^{996}$; when $m \leqslant 995$, $S(m) \geqslant 2 \cdot 10+10^{997}-10>10 \cdot 10^{996}$, so $S(996)$ is the smallest, i.e., placing a “+” between the 996th and 997th digits yields the smallest sum.
(2) When $R(m)=18 \cdot 10^{m}+10^{1992-m}$ is minimized, the product
$$
\left(2 \cdot 10^{m}-1\right)\left(10^{1992-m}-9\right)=2 \cdot 10^{1992}+9-18 \cdot 10^{m}-10^{1992-m}
$$

reaches its maximum. And $R(m)$ satisfies the relation:
$$
\begin{array}{l}
R(995)=18 \cdot 10^{995}+10^{997}=118 \cdot 10^{995}, \\
\text { when } m \geqslant 996 \text {, } R(m)>18 \cdot 10^{m} \geqslant 180 \cdot 10^{995}, \\
\text { when } m \leqslant 994 \text {, } R(m)>10^{1992-m} \geqslant 1000 \cdot 10^{995},
\end{array}
$$

so $R(995)$ is the smallest, i.e., placing a “$\times$” between the 995th and 996th digits yields the maximum product.