Combinatorial Maxima

Its not too tough to figure out that we need two cases here on the basis of parity,

Case-1 , when 'n' is odd,

$\frac{A_{n}}{A_{n-1}}=\frac{\binom{n}{\frac{n-1}{2}}}{\binom{n-1}{\frac{n-1}{2}}}=\frac{2n}{n+1}$

which gives n=19 as a solution

Case-2 , when 'n' is even,

$\frac{A_{n}}{A_{n-1}}=\frac{\binom{n}{\frac{n}{2}}}{\binom{n-1}{\frac{n-1}{2}}}=2$

which gives n={2,4,6,8,...,20}

So total of $\boxed{11}$ solutions!