APMO

Part 1: All polynomials with $\mathrm{deg}P=1$ satisfy the given property. Suppose $P(x)=cx+d$, and assume without loss of generality that $c>d\ge 0$. Denote $s_i=a_1+a_2+\cdots +a_i(\mathrm{mod}c)$. It suffices to show that there exist indices $i$ and $j$ such that $j-i\ge 2$ and $s_j-s_i\equiv d(\mathrm{mod}c)$.

Consider $c+1$ indices $e_1,e_2,\dots ,e_{c+1}>1$ such that $a_{e_l}\equiv d(\mathrm{mod}c)$. By the pigeonhole principle, among the $n+1$ pairs $\left(s_{e_1-1},s_{e_1}\right),\left(s_{e_2-1},s_{e_2}\right),\dots ,\left(s_{n+1-1},s_{n+1}\right)$, some two are equal, say $\left(s_{m-1},s_m\right)$ and $\left(s_{n-1},s_n\right)$. We can then take $i=m-1$ and $j=n$.

Part 2: All polynomials with $\mathrm{deg}P\ne 1$ do not satisfy the given property. Lemma: If $\mathrm{deg}P\ne 1$, then for any positive integers $A,B$, and $C$, there exists an integer $y$ with $|y|>C$ such that no value in the range of $P$ falls within the interval $[y-A,y+B]$.

Proof of Lemma: The claim is immediate when $P$ is constant or when $\mathrm{deg}P$ is even since $P$ is bounded from below. Let $P(x)=a_n{x}^n+\cdots +a_{1}x+a_0$ be of odd degree greater than 1, and assume without loss of generality that $a_n>0$. Since $P(x+1)-P(x)=a_{n}n{x}^{n-1}+\dots$, and $n-1>0$, the gap between $P(x)$ and $P(x+1)$ grows arbitrarily for large $x$. The claim follows. $\square$

Suppose $\mathrm{deg}P\ne 1$. We will inductively construct a sequence $\{a_i\}$ such that for any indices $i<j$ and any integer $k$ it holds that $a_i+a_{i+1}+\cdots +a_j\ne P(k)$. Suppose that we have constructed the sequence up to $a_i$, and $m$ is an integer with smallest magnitude yet to appear in the sequence. We will add two more terms to the sequence. Take $a_{i+2}=m$. Consider all the new sums of at least two consecutive terms; each of them contains $a_{i+1}$. Hence all such sums are in the interval $[a_{i+1}-A,a_{i+1}+B]$ for fixed constants $A,B$. The lemma allows us to choose $a_{i+1}$ so that all such sums avoid the range of $P$.

Alternate Solution for Part 1: Again, suppose $P(x)=cx+d$, and assume without loss of generality that $c>d\ge 0$. Let $S_i=\{a_j+a_{j+1}+\cdots +a_i(\mathrm{mod}c)\mid j=1,2,\dots ,i\}$. Then $S_{i+1}=\{s_i+a_{i+1}(\mathrm{mod}c)\mid s_i\in S_i\}\cup \{a_{i+1}(\mathrm{mod}c)\}$. Hence $|S_{i+1}|=|S_i|$ or $|S_{i+1}|=|S_i|+1$, with the former occurring exactly when $0\in S_i$. Since $|S_i|\le c$, the latter can only occur finitely many times, so there exists $I$ such that $0\in S_i$ for all $i\ge I$. Let $t>I$ be an index with $a_t\equiv d(\mathrm{mod}c)$. Then we can find a sum of at least two consecutive terms ending at $a_t$ and congruent to $d(\mathrm{mod}c)$.

Alternate Construction when $P(x)$ is constant or of even degree If $P(x)$ is of even degree, then $P$ is bounded from below or from above. In case $P$ is constant or bounded from above, then there exists a positive integer $c$ such that $P(x)<c$. Let $\{a_i\}$ be the sequence $$0,1,-1,2,3,-2,4,5,-3,\cdots$$ which is given by $a_{3n+1}=2n$, $a_{3n+2}=2n+1$, $a_{3n+3}=-(n+1)$ for all $n\ge 0$. Notice that for any $i<j$ we have $a_i+\cdots +a_j\ge 0$. Then for the sequence $\{b_n\}$ defined by $b_n=a_n+c$, clearly $b_i+\cdots +b_j\ge (a_i+\cdots +a_j)+2c>c$ which is outside the range of $P(x)$.

Now if $P$ is bounded from below, there exists a positive integer $c$ such that $P(x)>-c$. In this case, take $b_n=-a_n-c$. Then for all $i<j$ we have $b_i+\cdots +b_j\le -(a_i+\cdots +a_j)-2c<-c$ which is again outside the range of $P(x)$.