We will first define a function $f: \mathbb{N} \times \mathbb{Z}_{\ge 0} \to \{0, 1\}$, where $\mathbb{Z}_{\ge 0}$ indicates the non-negative integers, and then show that if the current player is faced with a particular pair of values $(m, n)$ then that player has a guaranteed win if $f(m, n) = 1$ but a guaranteed loss if $f(m, n) = 0$, both under the assumption that the winner makes the right moves. The definition will also indicate a winning strategy in all cases.

In our definition of $f$, we implicitly assume that $m, n$ are integers with $m > 0$ and $n \ge 0$ whenever we define $f(m, n)$. We first define $f(m, n)$ for $m = 1$:
$$f(1, n) := \begin{cases} 0, & n \text{ even,} \\ 1, & n \text{ odd.} \end{cases} \qquad (7)$$
We next define $f(m, n)$ for $m > 1$ and $n \le m$:
$$f(m, n) := \begin{cases} 0, & n < m - 1, \\ 1, & m - 1 \le n \le m. \end{cases} \qquad (8)$$
We call the values of $f$ defined above the *initial values of $f$*. The remaining values $f(m, n)$ are defined iteratively as follows, where we implicitly assume that $n > m > 1$.
$$f(m, n) := \begin{cases} 0, & \text{if } f(k, n-k) = 1 \text{ for } k = m \text{ and } k = m-1 \\ 1, & \text{otherwise.} \end{cases} \qquad (9)$$
We claim that the above partial definitions give a consistent definition of $f$ on its full domain. First note that (7), (8), and (9) apply to non-overlapping sets of values $(n, m)$, so they are clearly consistent with each other, and it is similarly clear that each is internally consistent.
It remains to check that they provide a consistent definition $f(m, n)$ for each allowable $m, n$. This is clear for $m = 1$ and for $n \le m$, so suppose $n > m > 1$. Assume that $f(i, j)$ is already consistently defined if either $i < m$ or $i = m$ and $j < n$. Equation (9) then defines $f(m, n)$ in terms of $f(m, n-m)$ and $f(m-1, n-m+1)$. Thus, the definition is extended consistently to $f(m, n)$. By induction on $n$, we have defined $f(m, n)$ for all $n \in \mathbb{Z}_{\ge 0}$, and now by induction on $m$, we have defined $f(m, n)$ for all allowable values of $(m, n)$. Next, we show that $f$ has the desired interpretation. First, (7) reflects the fact that with $m = 1$, the game is predetermined because there is no choice: players take turns removing one counter after another and parity shows that the winner corresponds to the values of $f(1, n)$.

Finally, we show inductively that (9) correctly indicates the winner in the remaining cases. Assuming that $f(i, j)$ correctly indicates the winner both for $i < m$ and for $i = m$ when $j < n$, then $f(k, n-k) = 1$ for $k \in \{m, m-1\}$ indicates that both of these are winning positions so, faced with initial parameters $(m, n)$, the current player is forced to hand a winning position to the opponent.

On the other hand, if $f(k, n - k) = 0$ for some $k \in \{m, m - 1\}$, then the player can remove $k$ counters to hand a losing position to the other player. (If both values are 0, then either move works.) We have shown that the values of $f(m, n)$ indicate the winner and provide a winning strategy.

Let us note two implications that follow from (9) and which we will use repeatedly:
$$\begin{aligned} f(m, n - m) &= 1 &\Rightarrow f(m, n) &= 1 - f(m - 1, n - m + 1), \\ f(m - 1, n - m + 1) &= 1 &\Rightarrow f(m, n) &= 1 - f(m, n - m). \end{aligned}$$
Let us show that
$$f(2, n) = \begin{cases} 0, & n \text{ divisible by } 4, \\ 1, & \text{otherwise.} \end{cases} \qquad (10)$$
In fact, if $n$ is odd, then $f(2, n) = 1$ because $f(1, n - 1) = 0$ by (7). If $n > 0$ is even, then $f(1, n - 1) = 1$, so $f(2, n) = 1 - f(2, n - 2)$. In view of the initial values $f(2, 0) = 0$ and $f(2, 2) = 1$, we get the desired values of $f$. We next show that
$$f(3, n) = \begin{cases} 1, & n = 3 \text{ or } (n > 0 \text{ and } n \text{ even}), \\ 0, & \text{otherwise.} \end{cases} \qquad (11)$$
The initial values give us (11) for $n \le 3$, so we suppose inductively that the equation in (11) holds as long as $n < 4k$ for some $k \in \mathbb{N}$. Since $f(2, 4k) = 0$, we have $f(3, 4k + 2) = 1$. As for $f(3, n)$ for $n \in \{4k, 4k + 1, 4k + 3\}$, in each of these cases $f(2, n - 2) = 1$ and so $f(3, n) = 1 - f(3, n - 3)$.
For $n = 4k$, we inductively have $f(3, n - 3) = f(3, 4n - 3) = 0$, so $f(3, 4k) = 1 - 0 = 1$. This in turn implies that $f(3, 4k + 3) = 1 - 1 = 0$. Similarly, $f(3, 4k + 1 - 3) = 1$, so $f(3, 4k + 1) = 1 - 1 = 0$. This completes the inductive step (from $n < 4k$ to $n < 4(k + 1)$), so (11) is fully valid.
We next show that
$$f(4, n) = \begin{cases} 0, & n < 3 \text{ or } n \equiv \pm 1 \pmod 8, \\ 1, & \text{otherwise.} \end{cases} \qquad (12)$$
The initial values give us (12) for $n \le 4$, so we suppose that $n > 4$. We get $f(4, 6) = 1$ because $f(4, 2) = 0$. For even $n > 6$, we get $f(4, n) = 1$ because $f(3, n - 3) = 0$.

Suppose therefore that $n > 4$ is odd. Since $f(3, n-3) = 1$, we have $f(4, n) = 1 - f(4, n-4)$. This equation plus the above initial values give us $f(4, k)$ for all odd $k > 4$.

We now have useful formulae for $f(m, n)$ for all $1 \le m \le 4$ and $n \ge 0$. For Alice to have a guaranteed win, she must choose the value of $n \ge 2024$ so that $f(m, n) = 1$ for each $m \in \{2, 3, 4\}$. By our above formulae, $m = 2$ rules out $n = 2024$, while $m = 3$ rules out $n = 2025$. However, $n = 2026$ has the desired properties, so that is our answer. More generally, for $n > 3$ we can say that $f(2, n) = f(3, n) = f(4, n) = 1 \pmod{4}$.