16. Several workers load and unload a batch of goods, with each worker having the same loading and unloading speed. If these workers work simultaneously, it will take $10 \mathrm{~h}$ to complete the loading and unloading. Now, the loading and unloading method is changed, starting with one person working, and then adding one more person every $t \mathrm{~h} (t$ is an integer). Each person who participates in the loading and unloading continues working until the task is completed, and the last person added works for a time that is $\frac{1}{4}$ of the time the first person works. Find:
(1) According to the changed loading and unloading method, how long does it take from start to finish?
(2) How many workers participate in the loading and unloading?
16. (1) Let the loading and unloading work take $x \mathrm{~h}$ to complete, then the first person worked for $x \mathrm{~h}$, and the last person worked for $\frac{x}{4} \mathrm{~h}$. The total work time of the two people is $\left(x+\frac{x}{4}\right) \mathrm{h}$, and the average work time per person is $\frac{1}{2}\left(x+\frac{x}{4}\right) \mathrm{h}$. According to the problem, the second person and the second-to-last person, the third person and the third-to-last person, etc., also have an average work time of $\frac{1}{2}\left(x+\frac{x}{4}\right) \mathrm{h}$. From the given conditions, we have
$$
\frac{1}{2}\left(x+\frac{x}{4}\right)=10 \text {. Solving for } x \text {, we get } x=16 \text {. }
$$
(2) Let there be $y$ people participating in the loading and unloading work. Since one person is added every $t \mathrm{~h}$, the last person works $(y-1) t \mathrm{~h}$ less than the first person. According to the problem, we have
$$
16-(y-1) t=16 \times \frac{1}{4},
$$

which simplifies to $(y-1) t=12$.
Solving this indeterminate equation, we get
$$
\begin{array}{c}
\left\{\begin{array} { l } 
{ y = 2 , } \\
{ t = 1 2 ; }
\end{array} \quad \left\{\begin{array} { l } 
{ y = 3 , } \\
{ t = 6 ; }
\end{array} \quad \left\{\begin{array}{l}
y=4, \\
t=4 ;
\end{array}\right.\right.\right. \\
\left\{\begin{array} { l } 
{ y = 5 , } \\
{ t = 3 ; }
\end{array} \quad \left\{\begin{array} { l } 
{ y = 7 , } \\
{ t = 2 ; }
\end{array} \quad \left\{\begin{array}{l}
y=13, \\
t=1 .
\end{array}\right.\right.\right.
\end{array}
$$

Thus, the number of people participating is $y=2$ or 3 or 4 or 5 or 7 or 13.