Note that $\frac{n}{k(n-k)}=\frac{1}{k}+\frac{1}{n-k}$. Therefore, multiplying the left side by $n$ yields $$\sum_{k=1}^{n-1}\frac{1}{k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}+\sum_{k=1}^{n-1}\frac{1}{n-k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}$$

These two terms are equal because replacing the index in the second term with $l=n-k$ yields the first one. This means $$\sum_{k=1}^{n-1}\frac{n}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1} =2\sum_{k=1}^{n-1}\frac{1}{k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1} $$

Hence, it is sufficient to show the following relation in order to prove the original claim. $$\sum_{k=1}^{n-1}\frac{1}{k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\frac{1}{2}\binom{2(n-1)}{n-1}$$

This relation can be proved using mathematical induction.

The relation clearly holds when $n=2$. Assuming it also holds when $n=m$, i.e.: $$\sum_{k=1}^{m-1}\frac{1}{k}\binom{2(k-1)}{k-1}\binom{2(m-k-1)}{m-k-1}=\frac{1}{2}\binom{2(m-1)}{m-1}$$

Then, when $n=m+1$, the left side becomes $$\sum_{k=1}^{m}\frac{1}{k}\binom{2(k-1)}{k-1}\binom{2(m-k)}{m-k} =\sum_{k=1}^{m-1}\frac{1}{k}\binom{2(k-1)}{k-1}\binom{2(m-k)}{m-k}+\frac{1}{m}\binom{2(m-1)}{m-1}$$

Note that $$\begin{align*} \binom{2(m-k)}{m-k} =\ &\frac{(2m-2k)!}{((m-k)!)^2} \\ =\ &\frac{(2m-2k)(2m-2k-1)}{(m-k)^2}\cdot\frac{(2m-2k-2)!}{((m-k-1)!)^2} \\ =\ &2\times\left(2-\frac{1}{m-k}\right)\binom{2(m-k-1)}{m-k-1} \end{align*}$$

Setting this to the previous relation gives $$\begin{align*} &\sum_{k=1}^{m}\frac{1}{k}\binom{2(k-1)}{k-1}\binom{2(m-k)}{m-k}\\ =\ & \sum_{k=1}^{m-1}\frac{2}{k}\left(2-\frac{1}{m-k}\binom{2(k-1)}{k-1}\binom{2(m-k-1)}{m-k-1}\right)+\frac{1}{m}\binom{2(m-1)}{m-1} \\ =\ &4\sum_{k=1}^{m-1}\frac{1}{k}\binom{2(k-1)}{k-1}\binom{2(m-k-1)}{m-k-1} - \\ & 2\sum_{k=1}^{m-1}\frac{1}{k(m-k)}\binom{2(k-1)}{k-1}\binom{2(m-k-1)}{m-k-1} + \frac{1}{m}\binom{2(m-1)}{m-1}\end{align*}$$

By the $2^{nd}$ equation given in this solution, the second term above equals $$\frac{4}{m}\sum_{k=1}^{m-1}\frac{1}{k}\binom{2(k-1)}{k-1}\binom{2(m-k-1)}{m-k-1}$$

Setting this back yields $$\begin{align*} &\sum_{k=1}^{m}\frac{1}{k}\binom{2(k-1)}{k-1}\binom{2(m-k)}{m-k} \\ =\ &4\left(1-\frac{1}{m}\right)\underbrace{\displaystyle\sum_{k=1}^{m-1}\frac{1}{k}\binom{2(k-1)}{k-1}\binom{2(m-k-1)}{m-k-1}}_{\scriptsize{apply\ the\ assumption}}  +\frac{1}{m}\binom{2(m-1)}{m-1}\\ =\ & 4\left(1-\frac{1}{m}\right)\cdot\frac{1}{2}\binom{2(m-1)}{m-1}+\frac{1}{m}\binom{2(m-1)}{m-1} \\ =\ &\left(2-\frac{1}{m}\right)\binom{2(m-1)}{m-1} \\ =\ &\frac{2m-1}{m}\frac{(2m-2)!}{((m-1)!)^2}\\=\ &\frac{1}{2}\cdot\frac{2m(2m-1)}{m^2}\cdot\frac{(2m-2)!}{((m-1)!)^2} \\=\ &\frac{1}{2}\binom{2m}{m}\end{align*}$$

This means that the relation holds for $n=m+1$. Thus, by the principle of mathematical induction, it holds for every $n\ge 2$.