[proofplan]
We apply the discrepancy estimate for the two endpoint intervals $[0,\eta)$ and $[1-\eta,1)$ to the finite sequence $P(1),\dots,P(N)$. The assumed inequality makes the resulting lower bound for the number of sampled fractional parts in those intervals strictly positive. Hence at least one sampled value lies within distance at most $\eta$ of an integer, and since $\eta<\delta$, this gives the desired $n$.
[/proofplan]
[step:Apply the discrepancy estimate to the endpoint intervals]
Define the real sequence $x_1,\dots,x_N$ by
\begin{align*}
x_n:=P(n)\qquad 1\le n\le N.
\end{align*}
Since $P\in\mathbb R[X]$, each $x_n$ is a real number. For a real number $t$, write $\{t\}:=t-\lfloor t\rfloor\in[0,1)$ for its fractional part. Since $0<\eta<\delta\le 1/2$, the parameter $\eta$ satisfies the hypothesis $0<\eta\le 1/2$ in [[Discrepancy Bound For Small Fractional Parts](/theorems/9095)][citetheorem:9095]. Applying that result to $x_1,\dots,x_N$ gives
\begin{align*}
\left|\#\{1\le n\le N:\{P(n)\}\in [0,\eta)\cup[1-\eta,1)\}-2\eta N\right|\le 2ND_N(P(1),\dots,P(N)).
\end{align*}
Therefore
\begin{align*}
\#\{1\le n\le N:\{P(n)\}\in [0,\eta)\cup[1-\eta,1)\}\ge 2\eta N-2ND_N(P(1),\dots,P(N)).
\end{align*}
[guided]
We want to force at least one of the fractional parts $\{P(1)\},\dots,\{P(N)\}$ to lie close to either endpoint $0$ or $1$ of the unit interval. Define
\begin{align*}
x_n:=P(n)\qquad 1\le n\le N.
\end{align*}
Because $P$ has real coefficients, each $x_n$ is a real number, so the finite-sequence discrepancy $D_N(x_1,\dots,x_N)$ is applicable. For each real number $t$, the notation $\{t\}:=t-\lfloor t\rfloor\in[0,1)$ denotes its fractional part.
The relevant target set is
\begin{align*}
[0,\eta)\cup[1-\eta,1).
\end{align*}
The theorem [Discrepancy Bound For Small Fractional Parts][citetheorem:9095] applies to [real numbers](/page/Real%20Numbers) $x_1,\dots,x_N$ and a parameter in the range $0<\eta\le 1/2$. We have $0<\eta<\delta\le 1/2$, so this hypothesis is satisfied. Applying the theorem with $x_n=P(n)$ gives
\begin{align*}
\left|\#\{1\le n\le N:\{P(n)\}\in [0,\eta)\cup[1-\eta,1)\}-2\eta N\right|\le 2ND_N(P(1),\dots,P(N)).
\end{align*}
Removing the absolute value in the lower-bound direction yields
\begin{align*}
\#\{1\le n\le N:\{P(n)\}\in [0,\eta)\cup[1-\eta,1)\}\ge 2\eta N-2ND_N(P(1),\dots,P(N)).
\end{align*}
This is the key quantitative point: discrepancy controls how far the actual count can fall below the expected length $2\eta$ times the sample size $N$.
[/guided]
[/step]
[step:Use the strict discrepancy hypothesis to obtain a nonempty set]
Let
\begin{align*}
A:=\{1\le n\le N:\{P(n)\}\in [0,\eta)\cup[1-\eta,1)\}.
\end{align*}
By the previous step,
\begin{align*}
|A|\ge N\left(2\eta-2D_N(P(1),\dots,P(N))\right).
\end{align*}
The hypothesis gives
\begin{align*}
2\eta-2D_N(P(1),\dots,P(N))>0.
\end{align*}
Since $N\in\mathbb N$ and $N\ge 1$, it follows that $|A|>0$. Thus $A$ is nonempty, and there exists $n\in\{1,\dots,N\}$ such that
\begin{align*}
\{P(n)\}\in [0,\eta)\cup[1-\eta,1).
\end{align*}
[guided]
Define
\begin{align*}
A:=\{1\le n\le N:\{P(n)\}\in [0,\eta)\cup[1-\eta,1)\}.
\end{align*}
The preceding discrepancy estimate gives
\begin{align*}
|A|\ge N\left(2\eta-2D_N(P(1),\dots,P(N))\right).
\end{align*}
The strict hypothesis says that the factor in parentheses is positive:
\begin{align*}
2\eta-2D_N(P(1),\dots,P(N))>0.
\end{align*}
Because the theorem assumes $N\ge 1$, multiplication by $N$ preserves strict positivity. Hence $|A|>0$, so $A$ is nonempty. Therefore there exists $n\in\{1,\dots,N\}$ with
\begin{align*}
\{P(n)\}\in [0,\eta)\cup[1-\eta,1).
\end{align*}
[/guided]
[/step]
[step:Convert endpoint membership into distance from an integer]
Choose $n\in A$. If $\{P(n)\}\in[0,\eta)$, then
\begin{align*}
\|P(n)\|_{\mathbb R/\mathbb Z}\le \{P(n)\}<\eta<\delta.
\end{align*}
If $\{P(n)\}\in[1-\eta,1)$, then
\begin{align*}
\|P(n)\|_{\mathbb R/\mathbb Z}\le 1-\{P(n)\}\le \eta<\delta.
\end{align*}
In either case,
\begin{align*}
\|P(n)\|_{\mathbb R/\mathbb Z}<\delta.
\end{align*}
This proves the existence of the required integer $n\in\{1,\dots,N\}$.
[guided]
Choose $n\in A$. By definition of $A$, the fractional part $\{P(n)\}$ lies in one of the two endpoint intervals. If $\{P(n)\}\in[0,\eta)$, then the nearest integer distance is bounded by the distance to $0\in\mathbb Z$, so
\begin{align*}
\|P(n)\|_{\mathbb R/\mathbb Z}\le \{P(n)\}<\eta<\delta.
\end{align*}
If $\{P(n)\}\in[1-\eta,1)$, then the nearest integer distance is bounded by the distance to $1\in\mathbb Z$, so
\begin{align*}
\|P(n)\|_{\mathbb R/\mathbb Z}\le 1-\{P(n)\}\le \eta<\delta.
\end{align*}
In both cases,
\begin{align*}
\|P(n)\|_{\mathbb R/\mathbb Z}<\delta.
\end{align*}
Thus the chosen integer $n\in\{1,\dots,N\}$ satisfies the theorem's conclusion.
[/guided]
[/step]
