We have \begin{align*} a-b &\equiv 62-75 \\ &\equiv -13 \\ &\equiv -13+99 \\ &\equiv 86\pmod{99}.\end{align*} This is not the answer since we want to find $ n $ with $ 1000\leq n<1099 $. Therefore we should add copies of 99 until we get into this range. Since 1000 is slightly more than $ 990=99\cdot10 $, we begin by adding 990. \[86\equiv 86+990\equiv1076\pmod{99}.\]That is in our range, so $ n=\boxed{1076} $.