Suppose that $p_1,...,p_k$ are distinct prime numbers.
Let $f(n,l)$ be equal to the number of elements from set $\{n+1,n+2,...,n+l\}$ that are divisible by some $p_1,...,p_k$. Is it true that $$\sup_{n,m,l \in \mathbb{N}} |f(n,l)-f(m,l)| = O(k).$$
Suppose that $p_1,...,p_k$ are distinct prime numbers.
Let $f(n,l)$ be equal to the number of elements from set $\{n+1,n+2,...,n+l\}$ that are divisible by some $p_1,...,p_k$. Is it true that $$\sup_{n,m,l \in \mathbb{N}} |f(n,l)-f(m,l)| = O(k).$$
No. There are various results which give counterexamples.
For example, Rankin's construction of large prime gaps boils down to the fact that if $p_1, \ldots, p_k$ denote all prime numbers below $x$, then there exists some $n$ with $f\left(n, \frac{x\log x\log\log\log x}{(\log\log x)^2}\right)=0$.
If $n=x, \ell=x^2-x$, then $f(n, \ell)=\pi(x^2)+\mathcal{O}(x)\sim\frac{x^2}{2\log x}$, but the average taken over all $n$ is $\ell \prod_{p\leq x}\left(1-\frac{1}{p}\right)\sim \frac{e^{-\gamma}x^2}{\log x}$.