Ireland

Let $S$ be the sum. Then $$1005-S=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots +\frac{1}{2011}=T.$$ Then $S$ is an integer if and only if $T$ is an integer. Let $M=3\cdot 5\cdot 7\cdot 9\dots 2009$. If $T$ is an integer then $MT$ is an integer. $$MT=\frac{M}{3}+\frac{M}{5}+\frac{M}{7}+\frac{M}{9}+\cdots +\frac{M}{2009}+\frac{M}{2011}.$$ Each term is an integer except the last one which is not since $2011$ is prime. Hence $T$ is not an integer.