UKMT Intermediate Challenge-III

If the divisor is not limited, then the largest remainder will $\left \lfloor \dfrac {2015} {2} \right \rfloor = 1007$ because $2015 = 1007+1008$ and $2015 \equiv 1007 \pmod{1008}$ . But $1008> 1000$ therefore the answer is unacceptable.

The next largest remainder is given by $\left \lfloor \dfrac {2015} {3} \right \rfloor = 671$ because $2015 = 671+2(672)$ and $2015 \equiv 671 \pmod{672}$ and $672<1000$ . Therefore the answer is $\boxed{671}$ .