Reposting my solution from yesterday: Suppose that $f$ has degree $d \ge 2$. Clearly, we may assume that the leading coefficient of $f$ is positive. In that case $f(n)+C \gg n^d$ for all $n$ where $C$ is a suitable constant depending only on $f$ and hence $$f(n+1)+\dots+f(n+k) +Ck \gg (n+1)^d+\dots+(n+k)^d \gg (n+1)^2+\dots+(n+k)^2 \gg kn^2+k^3.$$ So if we want to represent an integer $m \le M$, we need $kn^2+k^3 \ll M+k$ and hence $k \ll M^{1/3}$ and $n \ll M^{1/2}$. Hence, the number of choices of $(k,n)$ is only $\mathcal{O}(M^{5/6})$ so that almost all the integers up to $M$ are not represented, if $M$ is very large, in particular, there are infinitely many such numbers. (One can actually prove that the number of represented integers up to $M$ is $\mathcal{O}(M^{2/3})$ by counting slightly more carefully, but this is irrelevant here.)

Ok, in order to make this rigorous there are some annoying obstacles, so I prefer it as follows The senior coefficient of $f$ is positive. Let $N$ be sufficiently large positive integer. The plan is to count the possible integers in the interval $[N,2N]$ that can be represented as required. Note that $$f(n) \ge c_1n^2\qquad (1)$$ for sufficiently large integers $n$, where $c_1$ is a positive constant, that may depend on $f$. By $(1)$ it follows $$\#\{n\in\mathbb{N}: f(n)\le 2N \}\le c_2\sqrt{N}$$ where $c_2$ is a positive constant (possibly depending on $f$) Let us fix some $m\in \mathbb{N}$. The number of integers in $[N,2N$ that can be represented as sum of at most $m$ consecutive values of $f$ is at most $c_2m\sqrt{N}$. Let now estimate the number $N_m$ of integers $r\in[N,2N]$representable as $$r=f(n)+f(n+1)+\dots+ f(n+m-1)$$ It means $r\ge c_1mn^2$, thus $c_1mn^2\le 2N$ or $$n\le c_3\frac{\sqrt{N}}{\sqrt{m}}$$ $$n+m-1\le c_3\frac{\sqrt{N}}{\sqrt{m}}+m-1\le c_4\frac{\sqrt{N}}{\sqrt{m}}$$ which holds if $N$ is sufficiently large $N$. Further $$N_m\le \frac{c_4^2}{2}\left(\frac{\sqrt{N}}{\sqrt{m}}\right)^2=\frac{c_4^2}{2}\frac{N}{m}$$ Hence the number of integers in $[N,2N]$ that can be represented as required is at most $$c_2m\sqrt{N}+\frac{c_4^2}{2}\frac{N}{m}$$ which is less than $N$ for sufficiently large $m$ (and depending $N>m$).

In case you are referring to my leisure treatment of small values of $n$, I edited my solution slightly to address that. Otherwise, I don't see how it is not rigorous.

You have no problem with me! But note that many students do not even know even what "$\ll$" means.

The same method as the above ones can show that there are infinitely many primes not of the desired form, right? (Just replace $M$ by $M/\log M$, everywhere, having in mind the Prime Number Theorem, seems enough?) Is there are way to prove this in a purely number-theoretic fashion? (I have a good approach but with an awful gap, may post later.)

Let $g(x)=\sum\limits_{j=1}^x f(j)$. Note $g(x)$ is a polynomial of degree $\deg f+1$. Let $d=\deg f$ Say for sufficiently large $x$, $1-\epsilon < \frac{f(x)}{cx^d} < 1+\epsilon$, then $1-\epsilon < \frac{g(x)}{\frac{cx^{d+1}}{d+1}} < 1+\epsilon$ Notation. $a(x)=O(b(x))$ then $$|\lim_{x\to\infty} \frac{a(x)}{b(x)}|$$ is bounded. If $a(x)=o(b(x))$ this limit is 0. We group solutions into 2 exclusive groups that cover all possibilities: Group I: $g(a)<cn^{\frac{d+1}{d}-\delta}$ for a suitable $\delta<\frac{1}{d^2}$. This means $a,b=O(n^{\frac{1}{d}-\frac{\delta}{d+1}})$ so this produces $o(n^{\frac 2d})=o(n)$ solutions. Group II: $g(a)>cn^{\frac{d+1}{d}-\delta}$, then $f(a)>cn^{1-\frac{d \delta}{d+1}}$ so $a-b=O(n^{\frac{d \delta}{d+1}})$. For each choice of $a-b$, we are dealing with a polynomial $h(x)=g(x)-g(x-k)$, and each of them covers $O(n^{\frac 1d})$ numbers in $[1,n]$ so they cover at most $\frac 1c \cdot n^{\frac{d \delta}{d+1} + \frac 1d} = o(n)$ numbers. Adding two groups, the conclusion follows.

In the contest, I have reduced the question to the case of degree 2. Do you have any other specific solution for f, which has degree 2, in a straightforward way, without mentioning the aforementioned claims?

Is there any number-theoretical solution for the case when $f$ has integer coefficients?

Here is a short proof for $\deg f \geq 3$. We may assume that $f(x)$ has positive leading coefficient (otherwise it is bounded from above for $x>0$), thus $f(x) \to \infty$ as $x\to\infty$ and so the sets ${M\leq f(x)}$, $x=1,2,\ldots$ cover the positive integers. Now, for a fixed $x$, the number of solutions to $n+k \leq x$ (and hence the number of choices for $n$ and $k$) is at most $cx^2$ for some constant $c$; but $f(x) - cx^2 \to \infty$ for $\deg f \geq 3$ and so we are done.

I don't understand the symbols $O, \mathcal{O} and \ll$, what are they exactly? Can someone elaborate please? I couldn't find a solution sadly

Iora wrote: I don't understand the symbols $O, \mathcal{O} and \ll$, what are they exactly? Can someone elaborate please? I couldn't find a solution sadly All three are all synonyms for Big O notation.

Suppose not, that is, suppose that there are only finitely many positive integers which cannot be represented in the form $f(n+1)+f(n+2)+\cdots+f(n+k)$ for some positive integers $n$ and $k$. Let $S$ be the set of positive integers which cannot be represented in this form. Since $S$ is finite, there exists a positive integer $N$ such that $S\subseteq{1,2,\ldots,N}$. Consider the polynomial $g(x)=f(x+1)-f(x)$. Since $f$ has degree greater than 1, $g$ has degree at least 1. Thus, by the integer root theorem, there exists an integer $m$ such that $g(m)\neq 0$. Now consider the sum $$\begin{align*} \sum_{i=n+1}^{n+k}g(i)&=\sum_{i=n+1}^{n+k}[f(i+1)-f(i)] \ &=f(n+k+1)-f(n+1). \end{align*}$$ Thus, the set of possible values of $f(n+1)+f(n+2)+\cdots+f(n+k)$ is precisely the set of possible values of $f(n+1)+f(n+2)+\cdots+f(n+k+1)-f(n+1)$. Since $g(m)\neq 0$, there exist positive integers $n$ and $k$ such that $g(n+1)+g(n+2)+\cdots+g(n+k)\neq 0$. Then, the set of possible values of $f(n+1)+f(n+2)+\cdots+f(n+k)$ contains all the positive integers greater than or equal to $f(n+1)-\max{f(n+2),f(n+3),\ldots,f(n+k+1)}$. In particular, for any $i\in{1,2,\ldots,N}$, there exist positive integers $n$ and $k$ such that $f(n+1)+f(n+2)+\cdots+f(n+k)=i$. Thus, $i\notin S$, which contradicts the assumption that $S\subseteq{1,2,\ldots,N}$. Therefore, there must be infinitely many positive integers which cannot be represented in the form $f(n+1)+f(n+2)+\cdots+f(n+k)$ for any positive integers $n$ and $k$.

Assume the leading coefficient of $f$ is positive. Let $f_k(n)=f(n)+\cdots+f(n+k-1)$, so we want to show that there are infinitely many positive integers not representable as $f_k(n)$ for some $(k,n)$. Consider some large interval $[1,N]$ and suppose $f_k(n) \in [1,N]$. This clearly implies that $n$ is $O(N^{0.5})$. Furthermore, we can check (by estimating with an integral, for instance) that if $k~N^{0.4}$, then $f(1)~N^{1.2}$, hence $k$ must be $O(N^{0.4})$, so there are $O(N^{0.9})$ pairs $(k,n)$ that could possibly work. $\blacksquare$

Suppose that $f$ has degree $d \geq 2$. We may assume that the leading coefficient $a$ of $f$ is positive, otherwise $f(x)$ is bounded from above for $x>0$ and we are done. Let $C$ be a constant such that $f(x) \geq \frac{a}{2}x^d$ for all $x>C$. There are finitely many integers in the desired form with $n < C$ and $k < C$, so it suffices to show that there are finitely many with $n\geq C$ or $k \geq C$. Consider all integers in the interval $[1,M]$. In order for an integer in this interval to be representable in the desired form with corresponding $n$ and $k$, we must have $f(n+1) + f(n+2) + \cdots + f(n+k) \leq M$. On the other hand $$f(n+1)+\cdots+f(n+k) \geq \frac{a}{2}\left((n+1)^d+\cdots+(n+k)^d\right) \geq \frac{a}{2}\left((n+1)^2+\dots+(n+k)^2\right)$$ and hence $M \geq \frac{a}{2} \cdot kn^2$ and $M \geq \frac{a}{2} \cdot \frac{k^3}{3}$ (as $1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6} > \frac{k^3}{3}$). So if we want to represent an integer $m \leq M$, we need $k \leq AM^{1/3}$ and $n \leq AM^{1/2}$ for some constant $A$. Hence, the number of choices of $(k,n)$ is at most $A^2M^{5/6}$ and so at least $M - A^2M^{5/6} = M^{5/6}(M^{1/6} - A^2)$ integers are not representable in the desired form. As $M$ becomes very large, the latter expression grows arbitrarily large, so we are done.