Find all integer-coefficient polynomials $f(x)$ such that for all positive integers $n$, we have
$$
f(n) \mid 2^{n}-1 \text {. }
$$
Assume $f(x)$ is not a constant, without loss of generality, let the leading coefficient of $f(x)$ be positive, then there exists a positive integer $N$, such that when $x \geqslant N$, $f(x) \geqslant 2$.

Take any positive integer $n, n \geqslant N$, and take a prime factor $p$ of $f(n)$. Since $f(n) \mid 2^{n}-1$, we have $2^{n} \equiv 1(\bmod p)$.

Also, since $f(n+p) \equiv f(n) \equiv 0(\bmod p), f(n+p) \mid 2^{n+p}-1$, thus $2^{n+p} \equiv 1(\bmod p)$, hence $2^{p} \equiv 1(\bmod p)$.
By Fermat's Little Theorem, $2^{p} \equiv 2(\bmod p)$, thus $1 \equiv 2(\bmod p)$, a contradiction! Therefore, $f(x)$ is a constant, let $f(x)=a$.
Since $f(1) \mid 2^{1}-1$, i.e., $a \mid 1$, we know $a= \pm 1$. Therefore, the integer coefficient polynomials are
$$
f(x)=1 \text { and } f(x)=-1 .
$$