$$a_k:=\binom{n+\epsilon}{k}*\binom{2n-k}{n} =\frac{(n+\epsilon)!}{k!(n+\epsilon-k)!}\frac{(2n-k)!}{n!(n-k)!} $$
$$\frac{a_{k+1}}{a_k}=\frac{(n+1+\epsilon)!}{k!(n+1+\epsilon-k)!}\frac{(2n+2-k)!}{(n+1)!(n+1-k)!}\frac{k!(n+\epsilon-k)!}{(n+\epsilon)!}\frac{n!(n-k)!}{(2n-k)!} $$
$$\frac{a_{k+1}}{a_k}=\frac{(n+1+\epsilon)}{(n+1+\epsilon-k)}\frac{(2n+2-k)(2n+1-k)}{(n+1)(n+1-k)}$$
Now, solve
$$\frac{a_{k+1}}{a_k} \geq 1$$
This is quadratic in $k$ and gives you the monotony of $a_k$, which yields the max/min.