I think that every even target greater than $6$ is indeed a first player win. I have no proof but present evidence and speculation on what might lead to a proof.

For every even target at least up to $10000$ there is a strategy which is $2k \rightarrow 2k+1$ except for a small list of exceptions.
For example

- Target $T=9580$
Strategy: The other player will always state even numbers. Our response should be $2k \rightarrow 2k+1$ with these four exceptions $$4786 \rightarrow 7179,\ 7184 \rightarrow 7633,\ 9100 \rightarrow 9105,9572 \rightarrow 9574,\ 9578\rightarrow 9580$$

Let me justify that. Consider the sets $$W=\{9574,9580\}$$ $$L=\{4787,7185,9101,9573,9575,9579\}$$

These are the even numbers which it is a winner to state and the odd numbers which it is a loser to state. If the opponent never states a member of $W,$ they can't win. But those are $T=2^2\cdot 5\cdot 479$ and $9574=2\cdot 4787 .$
If we never state the odd numbers $9579,9575,9101,7185=T-1,T-5,T-479,T-2395$ or $4787=9574-4787,$ then they can't state anything from $W$ , unless perhaps we state an even number. We could state $9580$ and win. And if we state $9574=2\cdot 4787$ they can only reply $9576$ to which we can reply $9577 \notin L.$ So we just need responses *other than* $2k \rightarrow 2k+1$ for $2k+1 \in L.$ So preferably $2k \rightarrow 2k+t\lt T$ where $t \gt 1$ is an odd divisor of $k$ and $2k+t \notin L$. And we have done this for four of the five cases.For $9573-1=9572=2^2\cdot 2393$ there is no such $t$ to use. Fortunately, $9572 \rightarrow 9574$ allows us to state a winner.

**OPTIONAL DIGRESSION:** Here are three more examples

Since $p-1$ is not a power of $2,$ there is such an odd divisor.

- target $T=2p$ for $p$ a Fermat prime $p=2^s+1:$
Strategy: $2k \rightarrow 2k+1$ except $2p-2 \rightarrow 2p$ and $p-3 \rightarrow p-3+t$ where $t\gt 1$ is the next smallest odd divisor (so also $t \gt 3.$). here $W=\{p-1,2p\}\ \ L=\{p,2p-3\}.$

Since $p-1=2^s$ it is actually an even winner as the only responses are $p$ and even losers.

**END OF DIGRESSION**

Question which arise are:

How do we come up with the (or a) valid strategy?

How can we rule out $1 \in L?$

How large might the set of exceptions be?

The methods for an even target $T=2n$ is to start $W=\{2n\}$ and $L=\{\}$ where $W$ will be the even winners and $L$ will be the odd losers.

While there are unexamined members of $W \cup L$, consider the largest such.

If it is $2k \in W$, then add $2k-t$ to $L$ where $t$ ranges over the odd divisors of $k$. Then pick the next unexamined member.

If it is $2f+1 \in L$ then either find a good response to $2f$ or, if there is none, add $2f$ to $W$: Consider $2f+t$ where $t$ goes through the divisors of $2f$, As soon as you get an even winner in $W$ or and odd number not in $L$, add that to your list of exceptional rules. If none of those happen, add $2f$ to $W.$ Then pick the next unexamined member.

Stop when everything currently in $W \cup L$ has been examined.

The exceptions, given an even target $T \gt 6$, are $W,$ the even numbers it is a winner to state and $L$ the odd numbers it is a loser to state. For $\mathcal{T}$ a set of even targets (such as all even integers greater than $6$), One way to prove that every $T \in \mathcal{T}$ is a first player win is to show that every $T \in \mathcal{T}$ has smallest exception $e \gt 1$.

A stronger result might be easier. I would conjecture that $e \gt \frac{T}{4}.$

For even $6 \lt T \lt 10000$ there are $114$ cases where $\frac{e}{T} \lt \frac{15}{41}.$ Here are the first few pairs $[T,e]$ if we sort according to increasing ratio $\frac{e}{T}$

$$[114,31],[10,3],[30,9],[462,151],[130,43],[450,151],[1254,421],[2058,691],[1250,421],[3654,1231],[2050,691],[2530,853],[3650,1231],[4930,1663],[5730,1933],[8454,2851],[8450,2851],[9250,3121],[9570,3229]$$

So $\frac{e}{T} \gt \frac14$ (up to $T=10000$) and $\frac{e}{T} \gt \frac13$ for $T \gt 130.$ On the other hand, there are $T$ on the edge of the examined range with $\frac{e}{T} \lt 0.337$

There are $12$ cases for $4 \leq T \leq 10000$ with $2$ exceptions, The powers of $2.$

However there are $1285$ with exceptional set of size $3$, all cases of $T=2p$ and some, but not all, $T=2^sp$ for larger $s.$

Here are the first few counts: $$[2, 12], [3, 1285], [4, 28], [5, 609], [6, 187], [7, 60], [8, 525].$$
The largest are $72,75,80,83,84,94,97$ which occur once each.

Here is a cumulative plot

So just under $\frac23$ of the cases have exceptional sets of size $12$ or less and over $90\%$ have exceptional set of size $25$ or less.

You can just say that the target is the largest number which can be stated and the first player without a move loses. I didn't consider the case of first player without a move wins, but it should be largely similar.