Given $2 \leqslant r \leqslant t-2$, taking $i=2, t-1, t$ we get 
\[
\left\{\begin{array}{l}
(-1)^{2}\left(c_{2}-d_{2}\right)>0, \\ 
(-1)^{t-1}\left(c_{t-1}-d_{t-1}\right)>0, \\ 
(-1)^{t}\left(c_{t}-d_{t}\right)>0
\end{array} \Rightarrow \begin{array}{l}
c_{2}-d_{2}>0, \\ 
c_{t-1}-d_{t-1}>0, \\ 
c_{t}-d_{t}<0
\end{array} \Rightarrow \begin{array}{l}
c_{2}>d_{2}, \\ 
c_{t-1}>d_{t-1}
\end{array} \Rightarrow c_{2} \cdot c_{t-1}>d_{2} \cdot d_{t-1} \Rightarrow c_{t}>d_{t}\right.
\]
This contradicts $c_{t}-d_{t}<0$;
When $t$ is an odd prime, by the formula for positive divisors, $m, n$ are both powers of primes. 
Assume $m=p^{t-1}, n=q^{t-1}$, then $c_{i}=p^{i-1}, d_{i}=q^{i-1} \Rightarrow c_{i}-d_{i}=p^{i-1}-q^{i-1}$. In this case, the sign of $c_{i}-d_{i} (i=2,3, \cdots, t)$ is the same as the sign of $p-q$, which is a contradiction;
Therefore, $t$ is an odd composite number, so $t \geqslant 9$.
When $t=9$, construct $m=7^{2} \times 13^{2}, n=5^{2} \times 17^{2}$. It is easy to see that $c_{i}, d_{i}$ are respectively
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$i$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline$c_{i}$ & 1 & 7 & 13 & 49 & 91 & 169 & 637 & 1183 & 8281 \\
\hline$d_{i}$ & 1 & 5 & 17 & 25 & 85 & 289 & 425 & 1445 & 7225 \\
\hline$c_{i}-d_{i}$ & & + & - & + & + & - & + & - & + \\
\hline
\end{tabular}

In this case, $r=4$, so the smallest value of $t$ that satisfies the problem is 9.
12. Let $t$ be a positive integer greater than 3. Find the minimum value of $t$ such that there exist two positive integers $m, n$ satisfying: $m, n$ both have exactly $t$ positive divisors, and if the positive divisors of $m$ are arranged in ascending order as $c_{1}, c_{2}, \cdots, c_{t}$ (where $c_{1}=1, c_{t}=m$), and the positive divisors of $n$ are arranged in ascending order as $d_{1}, d_{2}, \cdots, d_{t}$ (where $d_{1}=1, d_{t}=n$), then there exists an integer $r(2 \leqslant r \leqslant t-2)$, such that
$$
(-1)^{i}\left(c_{i}-d_{i}\right)\left\{\begin{array}{l}
>0, i=2,3, \cdots, r \\
<0, i=r+1, \cdots, t
\end{array}\right.
$$