Forward equals backward

${n \choose 1234} = {n\choose 4321} \\ \dfrac{n!}{(n-1234)!1234!}=\dfrac{n!}{(n-4321)!4321!} \\ \dfrac{(n-4321)!4321!}{1234!(n-1234)!} = 1 \\ \Rightarrow n-4321=1234 \dots (\text{Since} 1234! \neq 4321! )\\ \Rightarrow \boxed{n=5555}$

Also there is a identity :

$\Large {n \choose a} = {n\choose n-a} \\ \Rightarrow n-1234=4321 \\ \Rightarrow \boxed{n=5555}$

I solved this in a sightly weird way

$\left( \begin{matrix} n \\ 1234 \end{matrix} \right) =\left( \begin{matrix} n \\ 4321 \end{matrix} \right) \\\frac { n! }{ 1234!(n-1234)! } =\frac { n! }{ 4321!(n-4321)! } \\ n!4321!(n-4321)!=n!1234!(n-1234)!\\ 4321!(n-4321)!=1234!(n-1234)!\\ \frac { 4321! }{ 1234! } =\frac { (n-1234)! }{ (n-4321)! } \\ (4321)(4320)\dots (1235)=(n-1234)(n-1235)\dots (n-4320)$

From here we notice that $4321=n-1234$ is the same as $4320=n-1235$ and so forth.

$4321=n-1234\\ n=\boxed { 5555 }$