The answer is $M^{\lfloor \frac n2 \rfloor},$ where $M=\prod\limits_{i=1}^r p_i$. Call such an operation a toggle at $a,b$. In this solution, $(a,b)$ denote their gcd and $[a,b]$ denote their lcm. CRUXThe main idea for Alice's strategy for $n$ odd, and Bob's strategy for $n$ even is to force the game to end with one number. The main idea for Alice's strategy for $n$ even and Bob's strategy for $n$ odd is to force the game to end with two numbers such that after the other player makes the last move, you still win. Alice's strategy for n oddAlice toggles $1, M^n$, setting $\alpha=1$. If Bob toggles $a,\frac{M^n}{a}$, then Alice takes $\alpha, [a,\frac{M^n}{a}]$. Since $\alpha\mid M^{\frac{n+1}{2}},$ this pair of moves simply remove $a,\frac{M^n}{a}$ (even if $\alpha=a$). If Bob toggles $a,b$ for $a,b\ne \alpha, ab\neq M^n$, then Alice toggles $\frac{M^n}{a}, \frac{M^n}{b}$. It can be easily shown that $[a,b] \cdot \gcd(\frac{M^n}{a}, \frac{M^n}{b})=M^n$. If Bob toggles $\alpha,b$, then Alice toggles $t,\frac{M^n}{t}$ and set $\alpha$ as their gcd. Note in this strategy, $\alpha\mid M^{\frac{n-1}{2}}$, and for all numbers $t$ other than $\alpha, M^n/\alpha$, the number of $t$'s on the blackboard is equal to the number of $\frac{M^n}{t}$'s on the blackboard. In the end, we arrive at $\alpha$ and win. Alice's strategy for n evenAgain, Alice toggles $(1,M^n)$ as her first move. Set $\alpha=1, \beta=M^{\frac n2}$. If Bob doesn't toggle $\beta$, Alice does exactly what she does if $n$ is odd, and the arguments would still work. If Bob toggles $\alpha, \beta$, let $\beta$ be their LCM and let $\alpha=\gcd(T,\frac{M^n}{T})$. The key thing to note is that $\alpha, \beta \mid M^{\frac n2}$. If Bob toggles $\beta,x$ for $x\neq \alpha$, then Alice toggles $[\beta,x], \frac{M^n}{x}$. This is valid because at any point, unless $x\in\{\alpha,\beta, \frac{M^n}{\alpha}, \frac{M^n}{\beta} \}$, we can see that the number of $x$ on the blackboard is the same as the number of $\frac{M^n}{x}$'s on the blackboard. Furthermore, the number of $\beta, \alpha$ is one more than number of $\frac{M^n}{\beta}, \frac{M^n}{\alpha}$, respectively. The numbers in the end of the process are $\alpha,\beta$, and their LCM is a divisor of $M^{\frac n2}$, as desired. Bob's strategy for n evenSet $\alpha=M^{\frac n2}$. If Alice toggles $x,y$, Bob toggles $\frac{M^n}{x}, \frac{M^n}{y}$. If Alice toggles $x,\frac{M^n}{x}$ for $x\neq\alpha, \frac{M^n}{\alpha}$, then Bob toggles $\alpha, \gcd(x,\frac{M^n}{x})$, and this new alpha is divisible by $M^{\frac n2}$. Notice the number of $\frac{M^n}{\alpha}$ is one less than the number of one less than the number of $\alpha$s, and for the other numbers, the number of $\frac{M^n}{x}$'s are equal to the number of $x$'s. In the end, we get $\alpha$, so we win. Bob's strategy for n oddSet $\alpha=M^{\frac{n-1}{2}}, \beta=M^{\frac{n+1}{2}}$. For other cases, refer to the strategy above. If Alice toggles $\beta$, $\alpha$, then Bob toggles an arbitary $T,\frac{M^n}{T}$, set $\beta\rightarrow (\beta,\alpha), \alpha\rightarrow [T,\frac{M^n}{T}]$, so we see that $M^{\frac{n-1}{2}}\mid \beta, M^{\frac{n+1}{2}}\mid \alpha$. If Alice toggles $\alpha,x$, then Bob toggles $(\alpha,x),\frac{M^n}{x}$ for $x\ne \frac{M^n}{\alpha}$, and reassigns $\alpha$ to that. The case for $\beta$ is similar. If $\alpha\beta=M^n$, then note the number of $t$ is exactly the number of $\frac{M^n}{t}$'s. Otherwise, note the number of $\alpha$, $\beta$, at the end of Bob's turn is one more than the number of $\frac{M^n}{\alpha}, \frac{M^n}{\beta}$, respectively. In the end, the last two numbers will be $\alpha, \beta$, so Bob achieves his goal.

The answer is $M^{\lfloor \frac n2 \rfloor},$ where $M=\prod\limits_{i=1}^r p_i$. Call such an operation a toggle at $a,b$. In this solution, $(a,b)$ denote their gcd and $[a,b]$ denote their lcm. CRUXThe main idea for Alice's strategy for $n$ odd, and Bob's strategy for $n$ even is to force the game to end with one number. The main idea for Alice's strategy for $n$ even and Bob's strategy for $n$ odd is to force the game to end with two numbers such that after the other player makes the last move, you still win. Alice's strategy for n oddAlice toggles $1, M^n$, setting $\alpha=1$. If Bob toggles $a,\frac{M^n}{a}$, then Alice takes $\alpha, [a,\frac{M^n}{a}]$. Since $\alpha\mid M^{\frac{n+1}{2}},$ this pair of moves simply remove $a,\frac{M^n}{a}$ (even if $\alpha=a$). If Bob toggles $a,b$ for $a,b\ne \alpha, ab\neq M^n$, then Alice toggles $\frac{M^n}{a}, \frac{M^n}{b}$. It can be easily shown that $[a,b] \cdot \gcd(\frac{M^n}{a}, \frac{M^n}{b})=M^n$. If Bob toggles $\alpha,b$, then Alice toggles $t,\frac{M^n}{t}$ and set $\alpha$ as their gcd. Note in this strategy, $\alpha\mid M^{\frac{n-1}{2}}$, and for all numbers $t$ other than $\alpha, M^n/\alpha$, the number of $t$'s on the blackboard is equal to the number of $\frac{M^n}{t}$'s on the blackboard. In the end, we arrive at $\alpha$ and win. Alice's strategy for n evenAgain, Alice toggles $(1,M^n)$ as her first move. Set $\alpha=1, \beta=M^{\frac n2}$. If Bob doesn't toggle $\beta$, Alice does exactly what she does if $n$ is odd, and the arguments would still work. If Bob toggles $\alpha, \beta$, let $\beta$ be their LCM and let $\alpha=\gcd(T,\frac{M^n}{T})$. The key thing to note is that $\alpha, \beta \mid M^{\frac n2}$. If Bob toggles $\beta,x$ for $x\neq \alpha$, then Alice toggles $[\beta,x], \frac{M^n}{x}$. This is valid because at any point, unless $x\in\{\alpha,\beta, \frac{M^n}{\alpha}, \frac{M^n}{\beta} \}$, we can see that the number of $x$ on the blackboard is the same as the number of $\frac{M^n}{x}$'s on the blackboard. Furthermore, the number of $\beta, \alpha$ is one more than number of $\frac{M^n}{\beta}, \frac{M^n}{\alpha}$, respectively. The numbers in the end of the process are $\alpha,\beta$, and their LCM is a divisor of $M^{\frac n2}$, as desired. Bob's strategy for n evenSet $\alpha=M^{\frac n2}$. If Alice toggles $x,y$, Bob toggles $\frac{M^n}{x}, \frac{M^n}{y}$. If Alice toggles $x,\frac{M^n}{x}$ for $x\neq\alpha, \frac{M^n}{\alpha}$, then Bob toggles $\alpha, \gcd(x,\frac{M^n}{x})$, and this new alpha is divisible by $M^{\frac n2}$. Notice the number of $\frac{M^n}{\alpha}$ is one less than the number of one less than the number of $\alpha$s, and for the other numbers, the number of $\frac{M^n}{x}$'s are equal to the number of $x$'s. In the end, we get $\alpha$, so we win. Bob's strategy for n oddSet $\alpha=M^{\frac{n-1}{2}}, \beta=M^{\frac{n+1}{2}}$. For other cases, refer to the strategy above. If Alice toggles $\beta$, $\alpha$, then Bob toggles an arbitary $T,\frac{M^n}{T}$, set $\beta\rightarrow (\beta,\alpha), \alpha\rightarrow [T,\frac{M^n}{T}]$, so we see that $M^{\frac{n-1}{2}}\mid \beta, M^{\frac{n+1}{2}}\mid \alpha$. If Alice toggles $\alpha,x$, then Bob toggles $(\alpha,x),\frac{M^n}{x}$ for $x\ne \frac{M^n}{\alpha}$, and reassigns $\alpha$ to that. The case for $\beta$ is similar. If $\alpha\beta=M^n$, then note the number of $t$ is exactly the number of $\frac{M^n}{t}$'s. Otherwise, note the number of $\alpha$, $\beta$, at the end of Bob's turn is one more than the number of $\frac{M^n}{\alpha}, \frac{M^n}{\beta}$, respectively. In the end, the last two numbers will be $\alpha, \beta$, so Bob achieves his goal.