jmc

If $n$ leaves a remainder of 1 when divided by all of these numbers then $n-1$ is a multiple of all of these. We compute the LCM of these numbers as $$\begin{array}{rl}\text{lcm}(2,3,4,5,6,7,8,9)& =\text{lcm}(5,6,7,8,9)\\ & =\text{lcm}(5,7,8,9)\\ & =5\cdot 7\cdot 8\cdot 9\\ & =2520.\end{array}$$ The smallest $n>1$ that satisfies $2520\mid n-1$ is $n=\boxed{2521}$.