Basically what I have tried is:
Since for any tournament of order $n\leq2$, there exists a vertex $w$ such that $s(w)\geq \frac{1}{2}(n-1)$. and also a vertex $w_1$ such that $s^-(w_1)\geq \frac{1}{2}(n-1)$
$T_n$ has an underlying graph of $K_n$ which is $(n-1)$-connected. Thus $\exists$ at least $(n-1)$ distinct $u-v$ paths for any $u$ and $v$.
If I choose $w$ as $u$ and $w_1$ as $v$, the most I will get is $51$ distinct paths?
I'm stuck. How do I proceed with this question?