Let $C$ and $T$ be large real numbers.
Consider the inequality $$\sum_{\gamma \in \Gamma : |\gamma-T| \leq 1} 1 \leq C\log(3T)-1,$$ where $\Gamma$ is a finite set of real numbers.
Does this inequality imply that there exists a real number $T'$ such that \begin{gather} |T-T'| \leq 1, \\ |\gamma - T'| \geq \frac{1}{C\log(3T)} \quad \text{for all} \quad \gamma \in \Gamma. \end{gather}
I'm reading a paper where an assertion of this type is made, but I can't see why it is true.
edit: Changed 2 to 3. But I don't think it should make a difference...
Let $N=C\log(3T)$. Your first says there are at most $N-1$ elements of $\Gamma$ that are in the interval $[T-1,T+1]$, which has length $2$. You are asking to find $T'\in[T-1,T+1]$ that is far from all the elements of $\Gamma$. If you think of spreading out the points to minimize the distances where $T'$ can go, they would be at $T-1+\frac 1N,T-1+\frac 3N,T-1+\frac 5N \ldots T-1+\frac {2N-1}N=T+1-\frac 1N$ All the gaps are of size $\frac 2N$ and the spaces at the ends are $\frac 1N$. The points $T-1, T+1,$ and the midpoints of the intervals would all be candidates for $T'$. If you move any points closer together to prevent putting $T'$ in an interval, you open up some other interval.
There is one nit that makes this not work. You could squeeze the points together by a tiny bit to prevent putting $T'$ anywhere between the first and last points. Then there could be points in $\Gamma$ at $T-1-\epsilon$ and $T+1+\epsilon$ that don't count in the initial count because they are outside the interval but prevent us from putting $T'$ in the ends of the interval. We can fix that by either saying at the end for all $\gamma \in \Gamma \cap [T-1,T+1]$ or by making the denominator in the distance between $\gamma$ and $T'$ into $N+1$ instead of $N$ so the end intervals are of length $\frac 2{N+1}$ instead of $\frac 1N$