Final Round of National Olympiad

Answer: 14.

Note that the fourth powers of even numbers are divisible by 16 and the fourth powers of odd numbers are congruent to 1 modulo 16. As $2013\equiv 13(\mathrm{mod}16)$, the desired representation must contain at least 13 odd summands.

Suppose that no more summands are needed. As $7^4=2401>2013$, each summand must be $1^4=1$, $3^4=81$ or $5^4=625$. There can be at most 3 summands 625 since $4\cdot 625>2013$. Therefore the number of summands not divisible by 5 is at least 10. The fourth power of an integer not divisible by 5 is congruent to 1 modulo 5, whereas $2013\equiv 3(\mathrm{mod}5)$. Hence the number of summands not divisible by 5 must be at least 13. This shows that the representation contains only summands 1 and 81, but 13 such numbers sum up to at most $13\cdot 81$ which is less than 2013. Thus representations with 13 summands are impossible.

On the other hand, 14 fourth powers is enough as $6^4+5^4+3^4+11\cdot 1^4=2013$.