【Analysis】Let's assume the four input numbers are $\mathrm{a}<\mathrm{b}<\mathrm{c}<\mathrm{d}$. Since the final output is the same, let's assume this result is $\mathrm{m}$. If $\mathrm{m}$ is an even number, since $4 \mathrm{k}+1$ is odd, odd $\neq$ even, so the final input result must also be an even number, which is $2 \mathrm{~m}$. By analogy, the four numbers input by the people would all be $8 \mathrm{~m}$, which contradicts the fact that the input numbers are all different positive integers. Therefore, $\mathrm{m}$ is an odd number.

Then, there are 2 choices for the previous step: $2 \mathrm{~m}$ or $\frac{m-1}{4}$.
If the previous step is $2 \mathrm{~m}$, since $2 \mathrm{~m}$ is an even number, the process must be:
$$
8 \mathrm{~m} \rightarrow 4 \mathrm{~m} \rightarrow 2 \mathrm{~m} \rightarrow \mathrm{m}
$$

Next, we examine the parity of $\frac{m-1}{4}$.
Similarly, if $\frac{m-1}{4}$ is even, then the process can only be:
$$
\mathrm{m}-1 \rightarrow \frac{m-1}{2} \rightarrow \frac{m-1}{4} \rightarrow \mathrm{m}
$$

This results in only 2 different input values, which contradicts the fact that the input numbers are all different positive integers. Therefore, $\frac{m-1}{4}$ is also odd.
Then, there are 2 choices for the previous step: $\frac{m-1}{2}$ or $\frac{\frac{m-1}{4}-1}{4}=\frac{m-5}{16}$. Next, we examine the parity of $\frac{m-5}{16}$.
Similarly, $\frac{m-5}{16}$ is odd.
Therefore, the four processes are:
$$
\begin{aligned}
8 \mathrm{~m} \rightarrow 4 \mathrm{~m} \rightarrow 2 \mathrm{~m} \rightarrow \mathrm{m} \\
\mathrm{m}-1 \rightarrow \frac{m-1}{2} \rightarrow \frac{m-1}{4} \rightarrow \mathrm{m} \\
\frac{m-5}{8} \rightarrow \frac{m-5}{16} \rightarrow \frac{m-1}{4} \rightarrow \mathrm{m} \\
\frac{m-21}{64} \rightarrow \frac{m-5}{16} \rightarrow \frac{m-1}{4} \rightarrow \mathrm{m}
\end{aligned}
$$

Since these numbers are all positive integers, $64 \mid \mathrm{m}-21$, and $\frac{m-21}{64}$ is odd. To make $\mathrm{m}$ the smallest, then $\frac{m-21}{64}=1 \Rightarrow m=85$.
At this point, the four input numbers are: $680, 84, 10, 1$.
Therefore, the value input by Dad Sheep is 680.
13. The left figure below is a strange black box. This black box has one input port and one output port. When we input a number into the input port, a number result will be produced at the output port, following these rules:
(1) If the input is an odd number $\mathrm{k}$, the output is $4 \mathrm{k}+1$.
(2) If the input is an even number $\mathrm{k}$, the output is $\mathrm{k} \div 2$.

For example, if the input is the number 8, the output will be $8 \div 2=2$. If the input is the number 3, the output will be $3 \times 4+1=13$. Now, if 3 such black boxes are connected in series, as shown in the right figure below, the output of the first black box becomes the input of the second black box, and so on. For example, if the input number is 16, after the first black box, the result is 8, which becomes the input of the second black box. After the second black box, the result is 4, which becomes the input of the third black box. After the third black box, the result is 2, which is the final output. We can represent this process as $16 \rightarrow 8 \rightarrow 4 \rightarrow 2$.

Now, Meiyangyang, Xiyangyang, Lazy Sheep, and Father Sheep input different positive integers into this series of black boxes, with Father Sheep inputting the largest number. The 4 final output results are the same. When this output result is the smallest, what is the input value of Father Sheep?