Even with relaxing the conditions , the answer is no for $n=44,46$ and all even $94 \leq n \leq 200.$ I didn't check any further but I doubt it is true ever again. We will say that $b$ covers the prime $p$ if $p \mid b.$

To review, we want a set of $k$ integers,$A=\{a_1,\cdots a_k\}$ where $k=\lfloor \frac{\pi(n)}2 \rfloor$ so that every prime less than $n$ is covered by at least one of the $2k$ members of $B=\{a_1,a_1+1,\cdots,a_k,a_k+1\}.$

The task is the same for $n=2m-1$ and $n=2m$ so there is no loss in assuming that $n$ is even. Actually, we could also restrict to the case that $2n-1$ is prime. Otherwise it is the same set of $\pi(n-2)=\pi(n)$ primes and the task is the same.

As noted, if $m \lt p \lt 2m$ is prime then one of the chosen $a_i$ must be either $p$ or $p-1.$ The issue will be that there are only a relatively few more primes in $\{2 \cdots m\}$ compared to $\{m+1,\cdots 2m-1\}$ and this doesn't allow us enough flexibility to cover all the smaller primes.

So if $\pi(2m)-\pi(m)=j$ we have $2^j$ starts to $A$ to consider. To simplify that we relax the problem:

We will set $k=\lceil \frac{\pi(n)}2 \rceil.$ Also, in the event that we choose $a=p \gt m$ a prime, we will be allowed to use all of $p-1,p,p+1$ in our cover $B$.

Since we are allowed to, we always will, it can't hurt.

In the case $m=22,n=44$ there are $8$ primes up to $22$ and $6$ more primes after that, $$23, 29, 31, 37, 41, 43.$$ These are all but one of of the members of $A.$

So, with the relaxed conditions, $B=\{22,23,24,28,29,20,31,32,36,37,38,40,41,42,43,44\}$ along with a pair, or perhaps triple, of consecutive integers. But we still have $13$ and $17$ uncovered and no pair or triple can cover both.

For $n=46$ we have the same primes to consider so that is also a failure.

In the case above we had all but one prime covered. As $n$ increases, so does the discrepancy.

Even with relaxing the conditions , the answer is no for $n=44$ and most $94 \leq n \leq 200.$ I didn't check any further but I doubt it is true ever again. (I relaxed the condition even further than when I first wrote this and haven't rechecked.) We will say that $b$ covers the prime $p$ if $p \mid b.$

To review, we want a set of $k$ integers,$A=\{a_1,\cdots a_k\}$ where $k=\lfloor \frac{\pi(n)}2 \rfloor$ so that every prime less than $n$ is covered by at least one of the $2k$ members of $B=\{a_1,a_1+1,\cdots,a_k,a_k+1\}.$

The task is the same for $n=2m-1$ and $n=2m$ so there is no loss in assuming that $n$ is even. Actually, we could also restrict to the case that $2n-1$ is prime. Otherwise it is the same set of $\pi(n-2)=\pi(n)$ primes and the task is the same.

As noted, if $m \lt p \lt 2m$ is prime then one of the chosen $a_i$ must be either $p$ or $p-1.$ The issue will be that there are only a relatively few more primes in $\{2 \cdots m\}$ compared to $\{m+1,\cdots 2m-1\}$ and this doesn't allow us enough flexibility to cover all the smaller primes.

So if $\pi(2m)-\pi(m)=j$ we have $2^j$ starts to $A$ to consider. To simplify that we relax the problem:

We will set $k=\lceil \frac{\pi(n)}2 \rceil.$ We will build the sets $A$ and $B$ as follows: Start with a list $L$ of the primes in question in decreasing order and $A=B=\emptyset$ . Take, p, the first so far uncovered prime in the list, choose a multiple $a=cp\leq n$ to add to $A.$ Add $a-1,a,a+1$ to $B$ and then move down $L$ to the largest still uncovered prime. Repeat until either all the primes are covered or $A$ is too big.

Clearly this relaxed condition gives bigger sets $B$ and lets us cover anything we could cover in the original problem.

In the case $m=22,n=44$ there are $8$ primes up to $22$ and $6$ more primes after that, $$23, 29, 31, 37, 41, 43.$$ These are all but one of of the members of $A.$

So, with the relaxed conditions, we so far have $B=\{22,23,24,28,29,20,31,32,36,37,38,40,41,42,43,44\}$ and can pick one more member of $A.$ Next down is $19$ which is already covered. For $p=17$ we need to pick an $a \in\{17,34\}$ But then we have all $7$ members of $A$ and $13$ is still uncovered.

For $n=43,45,46$ we have the same primes to consider so those are also a failure.

In the case above we had all but one prime covered. As $n$ increases, so does the discrepancy.

Even with relaxing the conditions , the answer is no for $n=44,46$ and all even $94 \leq n \leq 200.$ I didn't check any further but I doubt it is true ever again. We will say that $b$ covers the prime $p$ if $p \mid b.$

To review, we want a set of $k$ integers,$A=\{a_1,\cdots a_k\}$ where $k=\lfloor \frac{\pi(n)}2 \rfloor$ so that every prime less than $n$ is covered by at least one of the $2k$ members of $B=\{a_1,a_1+1,\cdots,a_k,a_k+1\}.$

The task is the same for $n=2m-1$ and $n=2m$ so there is no loss in assuming that $n$ is even. Actually, we could also restrict to the case that $2n-1$ is prime. Otherwise it is the same set of $\pi(n-2)=\pi(n)$ primes and the task is the same.

As noted, if $m \lt p \lt 2m$ is prime then one of the chosen $a_i$ must be either $p$ or $p-1.$ The issue will be that there are only a relatively few more primes in $\{2 \cdots m\}$ compared to $\{m+1,\cdots 2m-1\}$ and this doesn't allow us enough flexibility to cover all the smaller primes.

So if $\pi(2m)-\pi(m)=j$ we have $2^j$ starts to $A$ to consider. To simplify that we relax the problem:

We will set $k=\lceil \frac{\pi(n)}2 \rceil.$ Also, in the event that we choose $a=p \geq m$ a prime, we will be allowed to use all of $p-1,p,p+1$ in our cover $B$.

Since we are allowed to, we always will, it can't hurt.

In the case $m=22,n=44$ there are $8$ primes up to $22$ and $6$ more primes after that, $$23, 29, 31, 37, 41, 43.$$ This is all but one of of the members of $A.$

So, with the relaxed conditions, $B=\{22,23,24,28,29,20,31,32,36,37,38,40,41,42,43,44\}$ along with a pair, or perhaps triple, of consecutive integers. But we still have $13$ and $17$ uncovered and no pair or triple can cover both.

For $n=46$ we have the same primes to consider so that is also a failure.

In the case above we had all but one prime covered. As $n$ increases, so does the discrepancy.

Even with relaxing the conditions , the answer is no for $n=44,46$ and all even $94 \leq n \leq 200.$ I didn't check any further but I doubt it is true ever again. We will say that $b$ covers the prime $p$ if $p \mid b.$

To review, we want a set of $k$ integers,$A=\{a_1,\cdots a_k\}$ where $k=\lfloor \frac{\pi(n)}2 \rfloor$ so that every prime less than $n$ is covered by at least one of the $2k$ members of $B=\{a_1,a_1+1,\cdots,a_k,a_k+1\}.$

The task is the same for $n=2m-1$ and $n=2m$ so there is no loss in assuming that $n$ is even. Actually, we could also restrict to the case that $2n-1$ is prime. Otherwise it is the same set of $\pi(n-2)=\pi(n)$ primes and the task is the same.

As noted, if $m \lt p \lt 2m$ is prime then one of the chosen $a_i$ must be either $p$ or $p-1.$ The issue will be that there are only a relatively few more primes in $\{2 \cdots m\}$ compared to $\{m+1,\cdots 2m-1\}$ and this doesn't allow us enough flexibility to cover all the smaller primes.

So if $\pi(2m)-\pi(m)=j$ we have $2^j$ starts to $A$ to consider. To simplify that we relax the problem:

We will set $k=\lceil \frac{\pi(n)}2 \rceil.$ Also, in the event that we choose $a=p \gt m$ a prime, we will be allowed to use all of $p-1,p,p+1$ in our cover $B$.

Since we are allowed to, we always will, it can't hurt.

In the case $m=22,n=44$ there are $8$ primes up to $22$ and $6$ more primes after that, $$23, 29, 31, 37, 41, 43.$$ These are all but one of of the members of $A.$

So, with the relaxed conditions, $B=\{22,23,24,28,29,20,31,32,36,37,38,40,41,42,43,44\}$ along with a pair, or perhaps triple, of consecutive integers. But we still have $13$ and $17$ uncovered and no pair or triple can cover both.

For $n=46$ we have the same primes to consider so that is also a failure.

In the case above we had all but one prime covered. As $n$ increases, so does the discrepancy.