Bulgarian Spring Tournament

a) From Fermat's theorem and $y^2\equiv 1(\mathrm{mod}67)\iff y\equiv \pm 1(\mathrm{mod}67)$ it follows that any student number gives a remainder of $0$, $1$ or $66$ when divided by $67$. Let us consider the numbers $12^{66k+1}$, where $k\in \mathbb{N}$. They are presented as the sum of $12$ student numbers. Furthermore, by Fermat's theorem $12^{66k+1}\equiv 12(\mathrm{mod}67)$. This shows that $b(12^{66k+1})=12$ for every $k\in \mathbb{N}$. Therefore, the set here is infinite.

b) For $P\in \mathbb{Z}[X]$ let us set $\Delta (P)(x)=P(x+1)-P(x)$. It is clear that if $P$ is of degree $d$ with leading coefficient $a$, then $\Delta (P)\in \mathbb{Z}[X]$ is a polynomial of degree $d-1$ with leading coefficient $ad$. Consider the series of polynomials $P_1(x)=x^{33}$, $P_{k+1}=\Delta (P_k)$ for $k\in \mathbb{N}$. It is easy to see by induction on $k$ that for each $x\in \mathbb{Z}$, $k\in \mathbb{N}$ the number $P_k(x)$ is a sum of $2^{k-1}$ student numbers. Furthermore, we have that $P_{33}(x)=33!x+b$ for some $b\in \mathbb{Z}$. Since $1$ and $-1$ are student numbers, each integer is the sum of at most $2^{32}+33!<12^{12^{12}}$ student numbers. Then the set here is empty, and therefore finite.