Example 2: Two people, A and B, play the following game: They take turns writing a natural number on the blackboard, with the requirement that the newly written number cannot be expressed as a non-negative integer linear combination of the numbers already on the blackboard. That is, after writing numbers \(a_{1}, a_{2}, \cdots, a_{n}\), one cannot write \(x_{1} a_{1}+x_{2} a_{2}+\cdots+x_{n} a_{n}\left(x_{i} \in \mathbf{N}\right)\). For example, if 3 and 5 are already written on the blackboard, one cannot write 8 (since \(8 = 3 + 5\)) or 9 (since \(9 = 3 \times 3\)). The rule is: the one who writes 1 loses the game. Suppose the initial numbers written on the blackboard are 5 and 6, and A starts. Question: Who has a winning strategy?
First, we prove a lemma.
Lemma: Let $a, b$ be coprime positive integers. Then $ab - a - b$ is the largest integer that cannot be expressed as $ax + by$ where $x, y \in \mathbb{N}$.

Proof: By the coprimality of $a, b$ and Bézout's identity, for any integer $n$, there exist integers $x, y$ such that $n = ax + by$.
$$
\begin{array}{l}
\text{By } ax + by = a(x + b) + b(y - a) \\
= a(x - b) + b(y + a),
\end{array}
$$

we can assume $0 \leq x \leq b-1$.
If $n > ab - a - b$, then
$$
\begin{array}{l}
by = n - ax > ab - a - b - a(b-1) = -b \\
\Rightarrow y > -1 \Rightarrow y \geq 0.
\end{array}
$$

Thus, any integer greater than $ab - a - b$ can be expressed as
$$
\begin{array}{l}
ax + by \quad (x, y \in \mathbb{N}).
\end{array}
$$
If $ab - a - b = ax + by \quad (x, y \in \mathbb{N})$, then
$$
\begin{array}{l}
ab = a(x + 1) + b(y + 1) \\
\Rightarrow b \mid (x + 1) \Rightarrow x + 1 \geq b.
\end{array}
$$

Similarly, $y + 1 \geq a$.
Then $ab = a(x + 1) + b(y + 1) \geq 2ab$, which is a contradiction.
Thus, the lemma is proved.
Returning to the original problem.
By the lemma, we know that any integer greater than $5 \times 6 - 5 - 6 = 19$ can be expressed as $5x + 6y \quad (x, y \in \mathbb{N})$, except for $1, 5, 6$. Among the numbers less than 19 that can be written as $5x + 6y$, we have
$$
\begin{aligned}
5 \times 2 & = 10, 5 + 6 = 11, 6 \times 2 = 12, \\
5 \times 3 & = 15, 5 \times 2 + 6 = 16, \\
5 & + 6 \times 2 = 17, 6 \times 3 = 18,
\end{aligned}
$$

leaving nine numbers, which are
$$
2, 3, 4, 7, 8, 9, 13, 14, 19.
$$

Therefore, if Player A starts by writing 19, the remaining eight numbers can be divided into four pairs: $(2, 3), (4, 7), (8, 9), (13, 14)$. When Player B writes one of these numbers, Player A writes the other number in the pair. After all eight numbers are written, it will be Player B's turn to write 1, so Player A wins. Thus, Player A has a winning strategy.