I liked this problem the most out of all 8 problems We claim that the answer is the product of prime not larger than $n$ It is easy to construct such $P$ First, if a prime $p$ such that $p^2 \mid A$ exists, $P(p) \equiv p \pmod {p^2}$ so $p^2 \mid A \mid p$, contradiction. Therefore, $A$ is square free. Consider one prime factor of $A$. For all elements $x$ in $\mathbb{Z}_p$, $P(x) \equiv 0$ holds in $\mathbb{Z}_p$. Assume that $p$ such that $n < p$ exists. $P(x)$ has at most $n < p$ root or all the coefficients are equivalent to $0$ in $\mathbb{Z}_p$. However, since the coefficient of the first term is $1$, this implies contradiction to the assumption that $n<p$. Hence, $A$ is at maximum the product of primes not larger than $n$.

Note: the below solution is wrong due to an incorrect understanding of condition. Solution: We claim that the answer is $n!$ . To prove this works, consider taking the polynomial $P(x)=x(x+1)\cdots (x+n-1)$. This works because $P(i)=i(i+1)\cdots (i+n-1)={{i+n-1}\choose {i-1}}\times n!$ is clearly an integer for all $i=1,2,\dots ,n$. Now we show that $A\leq n!$. It is enough to show that $A\mid n!$. Considering$\pmod{A}$, the first condition implies that $P(z)$ is a multiple of $A$ for all positive integers $z$. Take the $n$th finite difference of $P(x)$, $\Delta ^nP(x)$. Since $P(x)$ is monic and has degree $n$, it is well known that $\Delta ^nP(x)=n!$ for all integers $x$. Yet by the first condition we know that $A$ must be a factor of $\Delta ^nP(0)$, which is exactly $n!$ and the proof is complete.

Here, first term means linear term. Maybe his wording caused a misunderstanding. In other words, $P(x)$ has the following form: $$P(x) = c_nx^n + c_{n-1}x^{n-1}+\dots+c_2x^2+x \ \ (c_n \neq 0)$$

Cute. The answer is $A = \prod_{p \leq n} p$ Firstly $A$ must be square-free, else if $p^2 \mid A$ then $x=p$ fails. Assume if a prime $q > n$ divides $A$. then viewing in $\mathbb{F}_q$ gives that the coefficients are $0$ mod $q$ which is a clear contradiction.

You can construct such a polynomial by induction on$ A_{m}$, where$ A_{m+1}= A_{m}*p_{m}$($p_{m}$ is the m-th prime). $A{1}=2$. $P(x)= x^2+x $works. Then let $A_{m+1}=A_{m}q. Q(x) = -x\prod_{i \leq q} (x-i)$ we know that the factor of degree $1$ has reminder $1$ when divided by $q$. So let $H(x)=Q(x)-qkx$ such that $H(x)$ has the term of degree $1$ equal to $1$. Notice that $deg(H(x))=q$ which is ok (when constructing the polynomial for a specific n just add $Ax^n$ in case the polynomial doesn't have degree $n$ ) and for every $x$ between $1$ and $q, H(x)$ is divisible by $q$. Let $P(x)$ be the polynomial that works for $A_{m}$, now we can just find all the coefficients of the required polynomial $W(x)$ of degree $>1$ (with Chinese reminder theorem) by imposing that $W(x)$ must be equal to $P(x)$ when "watched" $mod A_{m}$ and to $Q(x)$ when watched $mod p$