Bulgarian Spring Tournament

a) From Fermat's theorem and $y^2 \equiv 1 \pmod{67} \Leftrightarrow y \equiv \pm 1 \pmod{67}$ 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 $12^{66k+1} \equiv 12 \pmod{67}$. 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.