Except for $n = 1, 2, 4, 6$, Lee always has a winning strategy.
For convenience, let $[n] = \{1, 2, \dots, n\}$. We first prove a lemma:
Lemma: If Lee first chooses $n$, and Alpha has already chosen his $k$-th number ($k \ge 2$), then Lee can always choose his $k$-th number.
Proof: Let $a_1 < a_2 < \dots < a_k$ be the first $k$ numbers chosen by Alpha; note that $a_k < n$. Since $a_{i+1} - a_i > 1$, for every $1 \le i \le k$ there must exist $b_i$ such that $a_i < b_i < a_{i+1}$, where $a_{k+1} := n$. Among these $k$ values $b_i$, Lee has chosen at most $k-1$ of them, so he can always choose one of them as his $k$-th number. Q.E.D.

Back to the original problem. Without loss of generality, we assume that Alpha's first move is a number not exceeding $\frac{n+1}{2}$, so Lee's first move is to choose $n$.

- We first prove that for all $n \ne 1, 2, 4, 6$, Lee has a winning strategy. This splits into two cases:

* $n$ is an odd number greater than or equal to 3: By the lemma, as long as Alpha can choose, Lee can also choose. Moreover, if the game is a draw, Alpha must choose $\frac{n+1}{2}$ pairwise non-consecutive numbers from $[n]$, and this is only possible if these are all the odd numbers in $[n]$; but Lee has already taken the odd number $n$ at the start, so Alpha will eventually run out of numbers to choose first. In other words, Lee wins.

* $n$ is an even number greater than or equal to 8: Similar to the previous case, the lemma tells us that Lee can never lose, and for the game to be a draw, Alpha must choose $\frac{n}{2}$ pairwise non-consecutive numbers from $[n-1]$, and this is only possible if these are all the odd numbers in $[n-1]$. Therefore, as long as Lee, on his second move, chooses any number from $\{1,3,...,n-3\}$ that Alpha has not yet chosen, Alpha must lose (note that at this point Alpha has only chosen two numbers, so Lee must have such a number available to choose. Since Lee's first move was to choose $n$, he cannot choose $n-1$.) In summary, Lee wins.

- We next prove that for all $n = 1,2,4,6$, Lee has no winning strategy. In fact, we will prove that the game must be a draw.

* $n = 1,2$ are obvious.

* When $n = 4$, it is easy to see that Alpha's first move must be to choose $1$, otherwise Lee chooses $4$ and wins. Next, Lee's first move must be to choose $4$, otherwise Alpha's second move is to choose $4$ and wins. Alpha must then choose $3$, Lee must choose $2$, and it is a draw.

* When $n = 6$, Alpha's first move is to choose $1$. First consider the two sets $\{3,5\}$ and $\{4,6\}$: if Lee chooses the smaller (larger) one from one of these sets, Alpha then chooses the larger (smaller) one from the other set. And if Lee chooses $2$, Alpha chooses the remaining number from the two sets above. From this we see that Alpha can force at least a draw, and by the lemma Lee can also force at least a draw, so the game must be a draw.