I claim the answer is $20!$. First, taking $n=20!$, we get immediately that $n^2\mid \textstyle P=\prod_{0\le j\le 20}(n+j)$. We now prove $n\le 20!$ is necessary. Let $(n+i)^2\mid P$ for some $0\le i\le 20$. Using the fact $n+j\equiv j-i\pmod{n+i}$ for any $i,j$, we thus get \[ n+i \mid \prod_{\substack{0\le j\le 20 \\ j\ne i}}(n+j) \Rightarrow n+i \mid \prod_{\substack{0\le j\le 20 \\ j\ne i}}(j-i). \]Now, $\textstyle \prod_{j\ne i}|j-i| = i!(20-i)!$. Using the fact $\textstyle \binom{20}{i}$ is increasing on $i\in\{0,\dots,10\}$ and decreasing on $i\in\{10,\dots,20\}$, we get $i!(20-i)!$ is maximized at $i\in\{0,20\}$, with a maximum value of $20!$. From here, we get $n+i\le 20!$, so $n\le 20!$ necessarily holds.

Make $\prod_{i=0}^{20}{n+i}\equiv 0\pmod{(n+20)^2} \Longrightarrow \prod_{i=0}^{19}{n+i}\equiv 0\pmod{(n+20)}$ to maximize the value. If $n+20$ has prime factors $d_1, d_2\ldots,d_k$, then we wish to maximize the exponents of the primes in the factorization of $n+20$. Any $d_i<20, i\in\{1,2\ldots{k}\}$ since $d_i\mid{n+20}$, and another $n+r$ where $r\in\{1,2\ldots,19\}$ has a factor of $d_i,$ so $20-d_i=r\Longrightarrow d_i<20$. Let $p<20$ be a prime. Note that $\max{\nu_p{(\prod_{i=0}^{19}{n+i})}}=\sum_{i=1}^{s}{\lfloor{\frac{20}{p^s}\rfloor}}$ where $p^s\leq20\leq{p^{s+1}}$. This can easily be derived if we let $p\mid{n+20}$ and find the powers of $p$ in the value $\prod_{i=0}^{19}{n+i}\equiv \prod_{i=1}^{20}{i} \pmod{p}$. Testing all $p<20$ and making $n+20$ divisible by all the primes under 20(to maximize it), the maximum possible value of $n=2^{18}\cdot3^8\cdot5^4\cdot7^2\cdot11\cdot13\cdot17\cdot19$.

Note that finding $\nu_p{(\prod_{i=1}^{20}{i})}$ only works since $\nu_p{(n+20)}\geq s \Longrightarrow n+20\equiv 0\pmod{p^s}$.

Clearly $n=20!$ works. Assume there exists an $n>20!$ that also works and let $i$ be such that $$(n+i)^2|(n)(n+1)\dots (n+20)$$Then we must have $$(n+i)\leq \prod _{0\leq i\neq j\leq 20}\gcd(n+i,n+j)\leq \prod_{0\leq i\neq j\leq 20}|i-j|\leq 20!$$a contradiction.