Note that $n$ is a Pisot–Vijayaraghavan number with minimal polynomial $P(x) = x - n$. It follows that the largest real root of $x^{100} P(x) - P^*(x)$ where $P^*(x) = - nx + 1$ is the reciprocal polynomial must be a Salem Number by virtue of $100$ being large. It is well known that the fractional parts of Salem numbers are dense in $[0, 1]$, we win. References to both of these facts (though the first is not explicit in bounding) can be found in Algebraic Numbers and Fourier Analysis 1963, page 30-33, Theorem IV & V. Sketch of Polynomials causing Salem Number from book: Apply Rouche's to $(1 + \varepsilon) x^{100} P + P^*(x)$ to get that all other roots are inside the unit circle. For the large root $r$ of $P$, we get that if $n$ is large then $R(r + \varepsilon) \cdot \varepsilon$ has constant sign when sufficiently small (make this explicit if you care). IVT then gives us the root with modulus more than one. Then by Vieta's on the constant term, the roots of this polynomial are $r$, its inverse, and the rest must all be on the unit circle. $\blacksquare$ Sketch of density from book: Let the roots on the unit circle be $\alpha_1, \dots, \alpha_k$ and their conjugates, let the other two roots be $r$ and $\frac{1}{r}$. Then note that $\alpha_1, \dots, \alpha_k$ are multiplicatively independent over $\mathbb{Q}$ by taking a Galois automorphism. By Newton's sums, we have that \[ r^n \equiv -\sum_{i=1}^k (\alpha_i^n + \overline{\alpha_i}^n) + \frac{1}{r^n} \pmod{1} \] Apply Kronecker's density theorem for a large enough $n$ to get $-\sum_{i=1}^k \alpha_i^n$ to be whatever you want. $\blacksquare$

We use $\alpha$ to denote that real root larger than 1. Let $p\left(x\right)=\frac{x^{101}-nx^{100}+nx-1}{x-1}$. The existence of $\alpha$ guarantees $n\geq 1$, and it is obvious that $\frac{p\left(x\right)}{x^{50}}=T_{50}\left(x+\frac{1}{x}\right)-\left(n-1\right)\left(\sum\limits_{i=1}^{49}T_{i}\left(x+\frac{1}{x}\right)+1\right)$，where $T_{n}$ are the Chebyshev polynomials of the second kind. Plug in $x+\frac{1}{x}=2\mathrm{cos}\frac{k\pi}{50},k=0,1,...,49$, an easy calculation shows that $\frac{p\left(x\right)}{x^{50}}$ alternates its sign on these points; thus, $\frac{p\left(x\right)}{x^{50}}$, as a polynomial of $x+\frac{1}{x}$, has at least 49 distinct real roots in $\left(-2,2\right)$, which corresponds to 98 distinct complex roots of $p\left(x\right)$ with modulus 1, and the rest 2 are $\alpha$ and $\frac{1}{\alpha}$. Consider the minimal polynomial $q\left(x\right)$ of $\alpha$ over $\mathbb{Z}\left[x\right]$, then $q\mid p$. Note that $\alpha+\frac{1}{\alpha}-1<\frac{1+\sum\limits_{i=1}^{50}\left(\alpha^{i}+\frac{1}{\alpha^i}\right)}{1+\sum\limits_{i=1}^{49}\left(\alpha^{i}+\frac{1}{\alpha^i}\right)}=n<\alpha$, so neither $\alpha+\frac{1}{\alpha}$ nor $\alpha$ is an integer. Thus, the root of $q$ are $\alpha,\frac{1}{\alpha},z_{1},...z_{k},\overline{z_{1}},...,\overline{z_{k}}$，$z_{j}$ with modulus 1, possibly without $\frac{1}{\alpha}$, and $k\geq 1$. Let $z_{j}=\mathrm{exp}\left(2\pi i w_{j}\right)$, we claim that the $w_{j}$s and $1$ are linearly independent on $\mathbb{Z}$. Assume FTSOC that $n_{0}+\sum\limits_{j=1}^{k}{n_{j}w_{j}}=0$, and WLOG $n_{1}\ne 0$, then $\prod_{j=1}^{k}z_{j}^{n_{j}}=1$. As $q$ is irreducible, its Galois group is transitive, i.e. there exists an automorphism $\varphi$ of the Galois group s.t. $\varphi\left(z_{1}\right)=\alpha$. There is no $j$ for which $\varphi\left(z_{j}\right)=\frac{1}{\alpha}$, otherwise $z_{j}=\frac{1}{z_{1}}$, contradiction. So $\prod_{j=1}^{k}\varphi\left(z_{j}\right)^{n_{j}}=1$, but $\left|\varphi\left(z_{j}\right)\right|=1$ and $\varphi\left(z_{1}\right)>1$, which is absurd. Note that $\alpha^{m}+\frac{1}{\alpha^m}+\sum\limits_{j=1}^{k}\left(z_{j}^m+\overline{z_{j}^m}\right)$ is an integer (possibly without the $\frac{1}{\alpha^m}$ term). As $\alpha>1$, the fractional part of $\alpha^{m}+2\sum\limits_{j=1}^{k}\mathrm{cos}2m\pi w_{j}$ tends to 0, and by Kronecker' theorem, the fractional part of $\sum\limits_{j=1}^{k}\mathrm{cos}2m\pi w_{j}$ is dense on $\left(0,1\right)$, this finishes the proof. Of course there are (elementary) alternatives to the Galois theory part, i.e., first prove that one of the $w_{j}$ is irrational, then mimic the proof of Kronecker's theorem, as suggested by Cheng Jiang (2022 IMO CHN 6).