I can't resist some comments. I find it highly unlikely that what you ask is possible. The question seems a bit odd in that it is so specific when more basic questions are open.
In the following some estimates are rough and corrections are welcome. Let $p_i$ be the $i$th prime and $P_t$ the product of the first $t$ primes. For example $p_{10000}=104729$ and $P_{10000} \approx e^{104392.2}.$ This illustrates the estimate $\ln{P_t} \approx p_t.$
Define $h(t)$ to be the smallest $N$ so that out of every $N$ consecutive integers at least one is relatively prime to $P_t$. Equivalently, it is the largest $N$ such that some $N-1$ consecutive integers each have a prime divisor from among the first $t$ primes. Equivalent to that is that $h(t)$ is the largest $N$ such that there is a choice of $t$ residues
$ k_{1t},\cdots ,k_{tt} $ with $\cup_1^t A_{it} \supset [1,N-1].$ where $A_{it}=k_{it}+np_i.$
According to this article $h(t) \gt (2e^{\gamma}+o(1))\frac{p_t\ln{p_t}\ln_3{p_t}}{ln_2^2{p_t}}$ where $\ln_j$ is the $j$-fold iterated logarithm. This lower bound seems reasonably close for $t \lt 50$ (the limit of knowledge at the time, and perhaps now). The best known upper bound is $h(t) \lt \ln^2(P_t).$ Using the estimate above that $\ln{P_t} \approx p_t$ we then get that $h(t) \lt p_t^2.$ I have seen it conjectured in a paper from 1976 that perhaps $h(t)=O(t^{1+\epsilon}).$ Note that $p_t \approx t\ln{t}=o(t^{1+\epsilon})$ I don't know how current that is. An off topic note is that sometimes one can find a longer interval of integers all enjoying a divisor from $p_1,\dots,p_{t-1},p_{t+1}$
With the notation above a weaker form of your question is
Is there an integer $M$ and system of residues $k_i$ such that for $A_i=k_i+np_i$ we have
$\cup A_{i}=\mathbb{N}$
With only finitely many exceptions, $k_i \gt \frac{p_i^2}{M}$
(I didn't get the motivation to require $p_i$ relatively prime to $M$. Was it just to forbid at least one prime?)
So the situation is that, as far as I know, no-one can say if $h(n) \gt \frac{p_t^2}{M}$ infinitely often, and there may even be reasons to doubt that. This is the case that you may completely change the residues each time you bring a new prime into play. In your case you want a single list of residues.Note that the choices which provide a good example for $h(n)$ may not be good for extending to $h(n+1)$.
$h(9)=39$ as shown by
$$2, 3, 2, 5, 2, 11, 2, 3, 2, 13, 2, 23, 2, 3, 2, 7, 2, 19, 2, 3, 2, 17, 2, 5, 2, 3, 2, 11, 2, 7, 2, 3, 2, 5, 2$$
$$ 13, 2, 3, 2, *, 2, *, 2, 3, 2, *, 2, *, 2, 3, 2, *, 2, 5, 2, 3, 2, 7, 2, *$$
This illustrates that $1+2n,2+3n,4+5n ,6+11n,10+13n,12+23n,16+7n,18+19n$ and $22+17n$ cover all the integers from $1$ to $67$ except for $40,42,46,48,52,60,66.$
The best we can do to extend this introducing the further primes $29,31,37,41,43$ is $h(10) \ge 42,h(11) \ge 46,h(12) \ge 48,h(13) \ge 52,h(14) \ge 60$ etc.The actual values are $46, 58, 66, 74, 90.$ But each requires a new selection of residues.
Best Answer
This is partially an answer to Chuck.
The denominator is divisible by all primes $p$ such that $\dfrac{n}{\log{n}} \ll p < n$, so the denominator grows exponentially in $n$. Furthermore, I believe (and have been too lazy and incompetent to check) that the sum is (in absolute value) $\gg n^{-t}$. This implies that the numerator also grows exponentially in $n$. So to me this shows that the problem should be easy; how can there ever not be a prime $> cn$ dividing somethig like $e^n$? This is espeically true for $c=1$, since if we know that the numerator is only divisible by primes smaller than $n$, we actually know that all the prime divisors of it must be $\ll \dfrac{n}{\log{n}}$. So, since Gerry pointed out that my metaconjecture has been refuted, I'd like to coin a new one: for $c=1$ the conjecture is provable by a simple counting argument.