31. When $n=p^{k}, p$ is a prime, the greatest common divisor is $p$; in other cases, it is $1$. For the case $n=p^{k}$, prove that $p \left\lvert\,\binom{ n}{l}\right., 1 \leqslant l \leqslant n-1$, and $p \|\binom{ n}{p^{k-1}}$. Otherwise, use proof by contradiction. If the greatest common divisor $d>1$, let $p \mid d$. Since $d \mid n$, we can set $n=p^{k} n_{1}, p \nmid n_{1}>1$, and prove that $p \nmid\binom{n}{p^{k}}$, leading to a contradiction.