Balkan Mathematical Olympiad competition problems

First, we will prove the following lemma:
Lemma. Let $f$ be a polynomial with rational coefficients such that $f(n)$ is an integer for any integer $n$. Then there exist integers $a_0, a_1, \dots, a_p$ such that $f(x) = \sum_{i=0}^{p} a_i \binom{x}{i}$.
Proof. Let us first prove that for any polynomial with rational coefficients $f$ there are rational numbers $a_0, a_1, \dots, a_p$ (where $p = \deg f(x)$) such that $f(x) = \sum_{i=0}^{p} a_i \binom{x}{i}$.
We will prove this by induction on $p = \deg f(x)$, with the base case $p = 0$ being clear. Assuming that $p \ge 1$ and that the result holds for polynomials of degree not exceeding $p-1$, consider a polynomial $f(x)$ of degree $p$. Then choose $a_p$ such that $f(x) - a_p \binom{x}{p}$ has degree not exceeding $p-1$ (namely, if $a$ is the leading coefficient of $f$, choose $a_p = a \cdot p!$). By the inductive hypothesis we can write
$$f(x) - a_p \binom{x}{p} = \sum_{i=0}^{p-1} a_i \binom{x}{i}$$
for some rational numbers $a_0, \dots, a_p$, and thus $f$ has the required form. Assuming that $f(n)$ is an integer for all integers $n$, then
$$a_1 = (a_0 + a_1) - a_0 = f(1) - f(0)$$
is an integer, as a difference of two integers. Clearly, $a_0 = f(0)$ is also an integer. Assuming that $a_0, \dots, a_{k-1}$ are integers for some $k \ge 2$, the relation
$$f(k) = a_0 \binom{k}{0} + a_1 \binom{k}{1} + \dots + a_{k-1} \binom{k}{k-1} + a_k$$
shows that $a_k$ is an integer, as well. Therefore $a_0, a_1, \dots, a_p$ are all integers and the proof of the lemma is complete.

Now let's return to the original problem. For all $n > k$, we have that
$$2025^n = (2024 + 1)^n = \sum_{i=0}^{n} \binom{n}{i} 2024^i \equiv \sum_{i=0}^{k-1} \binom{n}{i} 2024^i \pmod{11^k}.$$
Let $Q(n) = \sum_{i=0}^{k-1} \binom{n}{i} 2024^i$, which is a polynomial with rational coefficients of degree $k-1$ in $n$. Then the condition is equivalent to
$$P(n) + Q(n) \equiv 0 \pmod{11^k}$$
for all positive integers $n > k$.
Let $Q(x) = \sum_{i=0}^{k-1} q_i x^i$ and $P(x) = \sum_{i=0}^{d} p_i x^i$. If $d \ge k-1$, we can select $p_i = -q_i$ for all $0 \le d \le k-1$ and $p_i = 11^k$ for $i > k-1$ and the condition is satisfied. Obviously all $q_i$ are rational numbers, and hence $P$ is a polynomial with rational coefficients.
Let $p_i = \frac{a_i}{b_i}$ where $a_i, b_i$ are integers such that $\text{GCD}(a_i, b_i) = 1$, for all $i \in \{0, 1, \dots, k-1\}$ (or if $p_i = 0$ we set $a_i = 0, b_i = 1$). Notice that $\binom{x}{i} \cdot 2024^i = \frac{x(x-1)\dots(x-i+1)}{1 \cdot 2 \dots i} \cdot 2024^i$, and since

$v_{11}(i!) = \lfloor \frac{i}{11} \rfloor + \lfloor \frac{i}{11^2} \rfloor + \dots < i = v_{11}(2024^i)$, the denominator of this fraction (after reduction) is not divisible by $11$, for all $i \in \{0, 1, \dots, k-1\}$. Hence, none of the $b_i$'s is divisible by $11$.
Let $S = \text{LCM}(b_0, b_1, b_2, \dots, b_{k-1})$ and let $S_{inv}$ be an integer such that $S \cdot S_{inv} \equiv 1 \pmod{11^k}$ ($S_{inv}$ exists because $S$ is not divisible by $11$). Define $P'(x) = P(x) \cdot S \cdot S_{inv}$. Now, first notice that $P'(x)$ has integer coefficients ($P(x) \cdot S$ has integer coefficients, and multiplying it by $S_{inv}$ doesn't change that fact) and due to the construction of $S_{inv}$, for each positive integer $n > k$ we have that $P'(n) \equiv P(n) \pmod{11^k}$.
Thus, $P'$ also satisfies the condition and has integer coefficients. In conclusion, all $d \ge k-1$ satisfy the problem condition.

Now, suppose that $d < k-1$ and let
$$R(x) = P(x) + Q(x).$$
Then $\text{deg} R(x) = k-1$ and $\frac{R(n)}{11^k}$ is an integer for all positive integers $n > k$ (hence, in fact, for all positive integers due to the periodicity of $R$ modulo $11^k$). Let
$$T(x) = \frac{R(x)}{11^k} = \sum_{i=0}^{k-1} \frac{r_i}{11^k} x^i.$$
Then the leading coefficient of $T(x)$ is equal to
$$\frac{r_{k-1}}{11^k} = \frac{q_{k-1}}{11^k} = \frac{2024^{k-1}}{(k-1)!11^k}.$$
Also, using the lemma we can write
$$T(x) = \sum_{i=0}^{k-1} a_i \binom{x}{i}$$
with $a_i \in \mathbb{Z}$ for all $0 \le i \le k-1$. Comparing the leading coefficients we get that
$$\frac{2024^{k-1}}{(k-1)!11^k} = \frac{a_{k-1}}{(k-1)!}$$
which implies that
$$\frac{2024^{k-1}}{11^k} = a_{k-1} \in \mathbb{Z},$$
a contradiction. In conclusion, the desired positive integers $d$ are all such that $d \ge k-1$.