Is there a subset of the positive integers with this property?

Does there exist an infinite subset $S$ of the positive integers whose complement in the positive integers is also infinite, such that $S$ has no infinite arithmetic progressions, but has arbitrarily large arithmetic progressions, and also, there exists at least one $k\geq2$ such that $S$ does not have a maximal arithmetic progression of length $k$ ?

Consider the following set of positive integers $S= \{1, 2, 4, 6, 8, 11, 14, 17, 20,\dots\}$. Constructed by the following pseudocode when END = $\infty$:

S = []
i = 0 # iteration
j = 1 # step
n=1 # actual number
for (i=0; i<END, i++){
    for (k = 0; k<j; k++){
        S.append(n)
        n += j
    }
    j++
}

Essentially we're picking a number and we're adding an arithmetic progression of lenght 1 and step 1. Then we add the numbers to create a progression of step 2 and lenght 2 and so on.

Thus we have arbitrary long arithmetic progressions but we don't have an infinite one, as all of them will be of a fixed amount. Finally, the complement of S in the positive integers is obviously infinite, as $\forall x \in S, x>2$ we have $x+1 \notin S$, and so if S is infinite, his complement must be also infinite.

By construction, S has one maximal arithmetic progression of every length possible, thus, if we eliminate the terms correspondent to the specified k we search we have constructed an example of the set you searched.

In fact, is there an example where for every $k\geq2$ , $S$ does not have a maximal arithmetic progression of length $k$?

To awnser this second part we will follow a similar argument. Consider now $S=\{1, 2, 4, 5, 10, 11, ...\}$. We have constructed this by using the following pseudocode:

n = 1
list = []
for (i=0; i<END; i++) {
   list.append(n)
   n++
   list.append(n)
   n = n*2
}

Then $S$ has an infinite amount of 2-maximal arithmetic progressions.

Let's see that there are no bigger arithmetic progressions. If there were to exist a 3-arithmetic progression (even if it's not maximal) we would have $x \in S, \exists k \in \mathbb{Z}^+$ such that $x, x+k, x+2k \in S$.

For construction, $k$ cannot be 1, then let $k>1$. If $x\in S$ then $x+k \notin S$ if $k<x$ as the next number in $S$ bigger than $x+1$ is at least $2x$ (it may be $2(x+1)$). Then for us to have $x, x+k, x+2k \in S$, $k$ must satisfy $k\geq x, k\geq x+k$. Particularly, we'd need $k \geq x+k \implies 0 \geq x$ which is a contradiction.

Therefore, have no such thing as a 3-arithmetic progression, as we tried to prove.