what a problem!

The answer is $\lambda = 2n - 1$. We start with the following lemma. Lemma. Let \[ A = \prod_{i=0}^k \gcd(x+i, y), \quad B = \gcd(x(x+1)...(x+k), y) \]for any positive integers $x, y, k$, then we have \[ A \geq B \geq \frac{A}{k!}. \]For the first part, we have for each prime $p$, \[ v_p(A) = \sum_{i=0}^k \min\{v_p(x+i), v_p(y)\} \geq \min\left\{ \sum_{i=0}^k v_p(x+i), v_p(y) \right\} = v_p(B) \]and so $B \mid A \implies A \geq B$. For the second part, consider $0 \leq j \leq k$ for which the value of $v_p(x+j)$ is the largest (remember this idea as it will be used again). For each $i \neq j$, we have $v_p(x+i) = v_p((x+i) - (x+j)) = v_p(i-j)$ if $v_p(x+i) < v_p(x+j)$, but if $v_p(x+i) = v_p(x+j)$ then we still have $v_p(x+i) = v_p(i-j)$ since $|i-j| < p^{v_p(x+i) + 1}$ (because otherwise, between $x+i$ and $x+j$ there is a multiple of $p^{v_p(x+j) + 1}$, which is absurd). This means \begin{align*} v_p(A) &= \min\{v_p(x+j), v_p(y)\} + \sum_{i=1}^{k-j} \min\{v_p(i), v_p(y)\} + \sum_{i=1}^{j} \min\{v_p(i), v_p(y)\} \\ &\leq v_p(B) + v_p((k-j)!) + v_p((j)!) = v_p(B) + \sum_{i=1}^{\infty} \left\lfloor \frac{k-j}{p^i} \right\rfloor + \sum_{i=1}^{\infty} \left\lfloor \frac{j}{p^i} \right\rfloor \\ &\leq v_p(B) + \sum_{i=1}^{\infty} \left\lfloor \frac{(k-j) + j}{p^i} \right\rfloor = v_p(B) + v_p(k!) \end{align*}Hence, we get $A \mid k! B \implies B \geq \frac{A}{k!}$. $\square$ Now let $x - y = d$. We use this lemma to show that \begin{align*} \gcd\left( \prod_{i=0}^{n-1} x+i, \prod_{i=0}^{n-1} y+i \right) &\geq c_1 \prod_{i=0}^{n-1} \gcd(x(x+1)...(x+n-1), y+i) \\ &= c_1 \prod_{i=0}^{n-1} \gcd((d-i)(d-i+1)...(d+n-i-1), y+i) \\ &\geq c_2 \prod_{i=0}^{n-1} \gcd(d-i, y+i) \cdot \gcd(d-i+1, y+i) \cdot ... \cdot \gcd(d+n-i-1, y+i) \\ &\geq c_2 \gcd\left( d, \prod_{j=0}^{n-1} y+j \right) \cdot \prod_{i=1}^{n-1} \gcd\left( d+i, \prod_{j=0}^{n-j-1} y+j \right) \cdot \prod_{i=1}^{n-1} \gcd\left( d-i, \prod_{j=i}^{n-1} y+j \right) \end{align*}for some constants $c_1, c_2$. Let the final expression above be $c_2 S$. We will show that $\gcd\left( \prod_{i=0}^{n-1} x+i, \prod_{i=0}^{n-1} y+i \right) \leq S$. First note that \[ \gcd\left( \prod_{i=0}^{n-1} x+i, \prod_{i=0}^{n-1} y+i \right) \leq \prod_{i=0}^{n-1} \gcd(x(x+1)...(x+n-1), y+i) = \prod_{i=0}^{n-1} \gcd((d-i)(d-i+1)...(d+n-i-1), y+i) \]so it suffices to show for all primes $p$ that \[ \sum_{j=0}^{n-1} \min\left\{ \sum_{i=-j}^{n-i-1} v_p(d+i), v_p(y+j) \right\} \leq \min\left\{ v_p(d), \sum_{j=0}^{n-1} v_p(y+j) \right\} + \sum_{i=1}^{n-1} \min\left\{ v_p(d+i), \sum_{j=0}^{n-j-1} v_p(y+j) \right\} + \sum_{i=1}^{n-1} \min\left\{ v_p(d-i), \sum_{j=i}^{n-1} v_p(y+j) \right\} \]Take $j'$ such that $v_p(y+j')$ is largest. Note that $n! \mid (d-j)(d-j+1)...(d+n-j-1)$ (since $\binom{d-j}{n} \in \mathbb{Z}$), so for each $j \neq j'$ we have $v_p(y+j) = v_p(j-j') \leq v_p(n!) \leq \sum_{i=-j}^{n-j-1} v_p(d+i)$. If $v_p(y+j') \leq \sum_{i=-j'}^{n-j'-1} v_p(d+i)$, then $LHS = \sum_{j=0}^{n-1} v_p(y+j)$. Consider the smallest values of $k, l \geq 0$ such that $v_p(d+k) \geq \sum_{j=0}^{n-k-1} v_p(y+j)$ and $v_p(d-l) \geq \sum_{j=l}^{n-1} v_p(y+j)$. Then, \[ RHS \geq \sum_{0 \leq j \leq n-k-1 \vee l \leq j \leq n-1}^{n-k-1} v_p(y+j) + \sum_{i=-l+1}^{k-1} v_p(d+i). \]It suffices to show that \[ \sum_{i=-l+1}^{k-1} v_p(d+i) \geq \sum_{j=n-k}^{l-1} v_p(y+j). \]If $k+l \leq n$ then the summation on the right is nonexistent and we are done, so assume $k+l > n$. Take $j''$ such that $v_p(y+j'')$ is largest for $n-k\leq j'' \leq l-1$. Then, \[ \sum_{j=n-k}^{l-1} v_p(y+j) = v_p(y+j'') + v_p((l-1-j'')!) + v_p((j''-n+k)!) \leq \sum_{i=-j''}^{n-j''-1} v_p(d+i) + v_p((l-1-j'')!) + v_p((j''-n+k)!)) \]Notice that $-j'' \geq -l+1$ and $n-j''-1 \leq k-1$, so it remains to show \[ \sum_{i=-l+1}^{-j''-1} v_p(d+i) + \sum_{i=n-j''}^{k-1} v_p(d+i) \geq v_p((l-1-j'')!) + v_p((j''-n+k)!)) \]which is true since $(l-1-j'')! \mid \prod_{i=-l+1}^{-j''-1} d+i$ and $(j''-n+k)! \mid \prod_{i=n-j''}^{k-1} d+i$. If $v_p(y+j') > \sum_{i=-j'}^{n-j'-1} v_p(d+i)$, we consider two cases: if $j' \neq j''$ then $LHS \leq \sum_{j=0}^{n-1} v_p(y+j)$ and we proceed exactly the same way as we did. If $j' = j''$ then $LHS = \sum_{j=0}^{n-1} v_p(y+j) + \sum_{i=-j'}^{n-j'-1} v_p(d+i) - v_p(y+j')$ and we instead want to prove \[ \sum_{i=-l+1}^{k-1} v_p(d+i) \geq \sum_{j=n-k}^{j'-1} v_p(y+j) + \sum_{j=j'+1}^{l-1} v_p(y+j) + \sum_{i=-j'}^{n-j'-1} v_p(d+i) \]which is still equivalent to what we did before. Finally, we have shown that $c_2 S \leq GCD \leq S$, therefore \[ GCD \leq S \leq (d-n+1)(d-n+2)...(d)(d+1)...(d+n-1) \leq d^{2n-1} \]which proves that $\lambda = 2n-1$ works. It remains to prove that, for infinitely many positive integers $d$, there is $y$ so that $S \geq cd^{2n-1}$ for some constant $c \geq 0$ (this proves $\lambda \geq 2n-1$). We can take $d = (2n)!k$ where $k \in \mathbb{N}$ is not divisible by any prime $p \leq 2n$, which asserts that $v_p(d+i)$ is largest for $i = 0$ and $v_p(d) = v_p((2n)!)$ (for each $p \leq 2n$). Reusing our main idea, we have \[ v_p((d-n+1)(d-n+2)...(d+n-1)) = v_p((2n)!) + 2v_p((n-1)!) = \text{const} \]for primes $p \leq 2n$. Meanwhile, for $p > 2n$, notice that at most one of $d+i$ is divisible by $p$ ($-n+1 \leq i \leq n-1$): if $i \geq 0$ then take $y$ so that $v_p(d+i) = v_p(y)$, otherwise take $y$ so that $v_p(d+i) = v_p(y+n-1)$. This guarantees for each summand $v_p(\gcd(d+i, ...)) = v_p(d+i)$. Construct $y$ with CRT, and we get \[ v_p(S) = v_p((d-n+1)(d-n+2)...(d+n-1)) \text{ for } p > 2n \implies S \geq c(d-n+1)(d-n+2)...(d+n-1) \]for some constant $c$. Hence, we are done.

We can simplify the proof for the upper bound. The main lemma below can be also used for proving the lower bound. Let us prove that $\lambda\le 2n-1$. The main lemma is as follows. Lemma 1. Let $x,n$ be positive integers. Let $p$ is a prime. Let $a\in\{0,\ldots,n-1\}$ such that $v_p(x+a)=\max_{0\le i\le n-1}v_p(x+i)$. Then $$v_p\left(\prod_{i=0}^{n-1}(x+i)\right)=v_p(x+a)+v_p(a!)+v_p((n-1-a)!).$$ Proof. First pick an integer $i\in\{0,\ldots,n-1\}\setminus\{a\}$. We will prove that $v_p(x+i)=v_p(a-i)$. First, observe that $$v_p(a-i)=v_p((x+a)-(x+i))\ge\min(v_p(x+a),v_p(x+i))=v_p(x+i).$$Hence, it remains to prove that $v_p(a-i)\le v_p(x+i)$. Assume not; that is, $v_p(a-i)>v_p(x+i)$. Therefore $$v_p(x+a)=v_p((a-i)+(x+i))=v_p(x+i).$$It follows that $v_p(a-i)>v_p(x+i)=v_p(x+a)$. But then there exists a multiple of $p^{v_p(x+a)+1}$ between $x+a$ and $x+i$, contradiction. As a result, \begin{align*} v_p\left(\prod_{i=0}^{n-1}(x+i)\right)&=\sum_{i=0}^{n-1}v_p(x+i)\\ &=v_p(x+a)+\sum_{i=0}^{a-1}v_p(a-i)+\sum_{j=a+1}^{n-1}v_p(j-a)\\ &=v_p(x+a)+v_p(a!)+v_p((n-1-a)!) \end{align*}as desired. $\square$ Corollary 2. Let $x,n$ be positive integers. Let $p$ is a prime. Then $$v_p\left(\prod_{i=0}^{n-1}(x+i)\right)\le\max_{0\le i\le n-1}v_p(x+i)+v_p((n-1)!).$$Proof. Let $a\in\{0,\ldots,n-1\}$ such that $v_p(x+a)=\max_{0\le i\le n-1}v_p(x+i)$. From lemma 1, it suffices to prove that $v_p(a!)+v_p((n-1-a)!)\le v_p((n-1)!)$. This follows from the fact that $\frac{(n-1)!}{a!(n-1-a)!}=\binom{n-1}{a}$ is an integer. $\square$

Back to the main problem. Let $d=x-y$. Let $p$ be an arbitrary prime. Let $a,b\in\{0,\ldots,n-1\}$ such that $v_p(x+a)=\max_{0\le i\le n-1}v_p(x+i)$ and $v_p(y+b)=\max_{0\le j\le n-1}v_p(y+j)$. Let $t=a-b\le |n-1|$. Then by corollary 2 we have \begin{align*} \min\left(v_p\left(\prod_{i=0}^{n-1}(x+i)\right),v_p\left(\prod_{j=0}^{n-1}(y+j)\right)\right)&\le\min(v_p(x+a),v_p(y+b))+v_p((n-1)!)\\ &\le v_p(d+t)+v_p((n-1)!). \end{align*}Since the segments $[d-n+1,d-1]$ and $[d+1,d+n-1]$ are disjoint, at least one of them does not contain $d+t$. Hence, the set $\{d-n+1,\ldots,d+n-1\}\setminus\{d+t\}$ contains a block of $n-1$ consecutive integers, so $(n-1)!(d+t)\mid\prod_{k=1-n}^{n-1}(d+k)$. It follows that \begin{align*} \min\left(v_p\left(\prod_{i=0}^{n-1}(x+i)\right),v_p\left(\prod_{j=0}^{n-1}(y+j)\right)\right)&\le v_p((n-1)!(d+t))\\ &\le v_p\left(\prod_{k=1-n}^{n-1}(d+k)\right). \end{align*} Since $p$ was arbitrary, we conclude that $$\left.\gcd\left( \prod_{i=0}^{n-1} (x+i), \prod_{i=0}^{n-1} (y+i) \right)\right |\prod_{k=1-n}^{n-1}(d+k).$$ Since $d\ge n$, we obtain that $$\gcd\left( \prod_{i=0}^{n-1} (x+i), \prod_{i=0}^{n-1} (y+i) \right)\le\prod_{k=1-n}^{n-1}(d+k).$$Moreover, $$\prod_{k=1-n}^{n-1}(d+k)=d\prod_{i=1}^{n-1}(d^2-i^2)\le d^{2n-1}.$$ Therefore $$\gcd\left( \prod_{i=0}^{n-1} (x+i), \prod_{i=0}^{n-1} (y+i) \right)\le d^{2n-1}$$as desired.

The answer is $\boxed{\lambda=2n-1}$. We will first prove the upper bound, and then the lower bound by looking for equality in the already used ideas. Let $f(x, y)$ denote the gcd, and let $d=x-y$. The key is to consider the quantity $$0<P=P(d):=\prod_{i=1-n}^{n-1} (d+i)\leq d^{2n-1}.$$ Proof that $\lambda=2n-1$ works: The relevance of $P$ is justified in the following lemma. Lemma:It holds that $f(x, y)\mid P(d)$. Note that, if we have proven this, then it is clear that $\lambda=2n-1$ is valid. Proof of the lemma: For each prime $p$, let $$a_p:= \min \left(\max_{0\leq i \leq n-1}(\nu_p(x+i)), \max_{0\leq j\leq n-1}(\nu_p(y+j))\right).$$ By definition it is true that $x-y\in \pm \{0, 1, \cdots, n-1\} \pmod {p^{a_p}}$, and hence $p^{a_p}\mid P$. For primes $p>n$ it holds that $a_p=\nu_p(d)$, so the conclusion follows. The case $p<n$ remains. If $a_p\leq \log_{p}(2n-1)$, then it is not hard to see that $\nu_p(2n-1)!\geq \nu_p(f(x, y))$, and since $(2n-1)!\mid P$, the conclusion again follows. Else $p^{a_p}>2n-1$, so there is exactly one index $1-n\leq i \leq n-1$ such that $p^{a_p} \mid d+i$. Without loss of generality, let $a_p=\nu_p(x+k)$ for some index $k$. Then the conclusion follows by taking everything modulo $p^{a_p}$ to note that, in fact $$\nu_p(x(x+1)\cdots (x+n-1))\leq \nu_p(P).$$$\square$ Thus, this direction of the proof is concluded. $\blacksquare$ Proof that $\lambda=2n-1$ is minimal: The key is that the "converse" of the above lemma also holds. Lemma: Let $d=x-y$ be fixed, and consider a prime $p>2n-1$ such that $p^t\mid P(d)$. Then, we can choose $x, y \pmod{p^t}$ such that $p^t\mid f(x, y)$. Proof: This is quite clear. Since $p>2n-1$, there is exactly one index $i$ such that $p\mid d+i$, so we will in fact have $p^t\mid d+i$. If $i\geq 0$ we may choose $x\equiv -i \pmod{p^t}$, $y\equiv 0\pmod{p^t}$. If $i< 0$ we may choose $x\equiv 0 \pmod{p^t}$, $y\equiv i\pmod{p^t}$. $\square$ Hence, with a fixed $d$, by solving the CRT system for $x, y$ we can find $x, y$ such that $$f(x, y)\geq \prod_{p\mid P(d), \quad p>2n-1} p^{\max_{1-n\leq i \leq n-1}(\nu_p(d+i))}$$so it suffices to find values of $d$ such that the above product is of the order $O(d^{2n-1})$. We can do this by bounding the $\nu_p$ of primes $p<2n-1$: for every prime $p<2n-1$, we choose $d$ such that $$d-n+1\equiv 1 \pmod {p^{\lceil\log_p(2n-1)\rceil+100}}$$and solve for $x-y$ using CRT. Like so, we achieve that all $2n-1$ factors of $P$ have their $\nu_p$ bounded by $\lceil\log_p(2n-1)\rceil+100>\log_p(2n-1)$, so it holds that $$\prod_{p\mid P, \quad p>2n-1} p^{\max_{1-n\leq i \leq n-1}(\nu_p(d+i))}\geq \frac{\prod_{i=1-n}^{n-1} (d+i)}{\left(\prod_{p<n} p^{\log_p(2n-1)}\right)^{2n-1}} = O(d^{2n-1}),$$which is the desired conclusion. $\blacksquare$