Three, (50 points) Let the natural number $k$ satisfy $1 < k$, such that $a_{m}$ is less than at least $k-1$ of the numbers $a_{1}, a_{2}, \cdots, a_{k}$. It is known that the number of sequences satisfying $a_{m}=1$ is $\frac{100!}{4}$. Find the value of $k$.
Three, rearrange $a_{1}, a_{2}, \cdots, a_{k}$ into $b_{1}t(i=k+1, k+2, \cdots, m-1) \text {. }$
$$
\text{When } t \text{ is fixed, by } b_{1}t, \text{ there are } \mathrm{C}_{100-\text{ ways to choose. }}^{k-2}
$$
Arranging $b_{1}, b_{2}, \cdots, b_{k}$ has $k!$ ways. Thus, determining $a_{1}, a_{2}, \cdots, a_{k}$ has $(t-2) \mathrm{C}_{100-t}^{k-2} k!$ ways.

From the previous analysis, $a_{i}>t(i=k+1, k+2, \cdots, m-1)$, in the $100-t$ numbers greater than $t$, excluding $b_{3}$, $b_{4}, \cdots, b_{k}$, there are $100-t-(k-2)=102-t-k$ numbers left, hence there are $\mathrm{A}_{102-t-k}^{m-k-1}$ ways to choose.

Also, $a_{m}=1$ is fixed, and the remaining numbers $a_{m+1}, a_{m+2}, \cdots, a_{100}$ can be arranged in $(100-m)!$ ways.
In summary, the number of permutations satisfying $a_{m}=1$ is
$$
\begin{array}{l}
=\sum_{i=3}^{100-k}(t-2)!k(k-1)(100-t)!\mathrm{C}_{100-k}^{t-2} \\
\end{array}
$$
(Due to $\mathrm{C}_{n}^{m}=\mathrm{C}_{n-1}^{m}+\mathrm{C}_{n-1}^{m-1}$ )
$$
\begin{array}{l}
=\sum_{t=3}^{\text {KC- }-k} k(k-1)(100-t)!\frac{(100-k)!}{(102-k-t)!} \\
=k(k-1)(100-k)!\cdot \\
\sum_{t=3}^{100-k} \frac{(100-t)!(k-2)!}{(102-k-t)!(k-2)!} \\
=k!(100-k)!\sum_{t=3}^{100-k} \mathrm{C}_{100-1}^{k-2} \\
=k!(100-k)!\mathrm{C}_{98}^{k-1} \\
=k!(100-k)!\frac{98!}{(k-1)!(99-k)!} \\
=k(100-k) 98 \text { ! . } \\
\text { Given } T=\frac{100!}{4} \text {, we have } \\
k(100-k) 98!=\frac{100!}{4} \\
\Rightarrow k^{2}-100 k+99 \times 25=0 \\
\Rightarrow k=45 \text { or } 55 \text {. } \\
\end{array}
$$