Saudi Booklet

Let $f(x) = x^7 - 1$. We have $x^{7k+r} \equiv x^r \pmod{f(x)}$ for all $r = 0, 1, \ldots, 6$ and $k$ is a positive integer. We will group together all exponents divided by $7$ with remainder $r$. We have
$$\begin{aligned} P(x) &= a_{2025}x^{2025} + \cdots + a_2x^2 + a_1x + a_0 \\ &\equiv (a_0 + a_7 + \cdots + a_{2023}) + (a_1 + a_8 + \cdots + a_{2024})x + \cdots \\ &\quad + (a_6 + a_{13} + \cdots + a_{2022})x^6 \pmod{f(x)}. \end{aligned}$$
We see that the right-hand side is a polynomial of degree not exceeding $6$, so to have congruence $0$ modulo $f(x)$, we need the right-hand side to be identically $0$. From that, we will get
$$\sum_{i=0}^{289} a_{7i} = \sum_{i=0}^{289} a_{7i+1} = \sum_{i=0}^{289} a_{7i+2} = \sum_{i=0}^{288} a_{7i+3} = \sum_{i=0}^{288} a_{7i+4} = \sum_{i=0}^{288} a_{7i+5} = \sum_{i=0}^{288} a_{7i+6} = 0.$$
We see that in the first three sums, there are $290$ terms and in the remaining sums, there are $289$ terms. So in the first three sums, we choose $145$ number $1$ and $145$ number $-1$, and the remaining sums have at least one $0$ (otherwise the sum of $289$ odd numbers cannot be zero). Thus, there are at least $4$ coefficients equal to $0$, so there are at most $2026 - 4 = 2022$ non-zero coefficients.

To construct, we just choose the coefficients based on the above idea with $a_{2025} \neq 0$. The answer is $2022$.

$\boxed{2022}$