jmc

We have $$\begin{array}{rl}a-b& \equiv 62-75\\ & \equiv -13\\ & \equiv -13+99\\ & \equiv 86(\mathrm{mod}99).\end{array}$$This is not the answer since we want to find $n$ with $1000\le n<1099$. Therefore we should add copies of 99 until we get into this range. Since 1000 is slightly more than $990=99\cdot 10$, we begin by adding 990. $$86\equiv 86+990\equiv 1076(\mathrm{mod}99).$$That is in our range, so $n=\boxed{1076}$.