Let $i,j\in\{1,2,\dots,2019\}, i\ne j$, fixed. Consider $x_i, x_j$ variable and $x_k$ fixed, $\forall k\in\{1,2,\dots,2019\}, k\ne i, k\ne j$. Denote $a_i=x, a_j=y$. $\dfrac{1}{x+2019}+\dfrac{1}{y+2019}=\dfrac{1}{2019}-\sum_{k\in\{1,2,\dots,2019\}\setminus\{i,j\}}\dfrac{1}{a_k+2019}=E\in\left(0,\dfrac{1}{2019}\right)$. Let be $a\in\mathbb{R}$ such that $\dfrac{2}{a+2019}=E\Longrightarrow a=\dfrac{2}{E}-2019>2019$. We will determine the minimum value of $xy$. Denote $S=x+y, P=xy$. $\dfrac{1}{x+2019}+\dfrac{1}{y+2019}=E\Longrightarrow \dfrac{S+4038}{P+2019S+2019^2}=E\Longrightarrow$ $\Longrightarrow P=\left(\dfrac{1}{E}-2019\right)\cdot S+2019\cdot\left(\dfrac{2}{E}-2019\right)=\dfrac{a-2019}{2}\cdot S+2019\cdot a$. Using AM-GM: $S=x+y\ge 2\sqrt{xy}=2\sqrt{P}$. Results: $P\ge (a-2019)\sqrt{P}+2019a$. With the notation $t=\sqrt{P}$ results the inequality $t^2-(a-2019)t-2019a\ge 0$, with the positive solution $\sqrt{P}=t\ge a\Longrightarrow P\ge a^2$. The equality occurs for $S=2\sqrt{P}\Longrightarrow x=y=a$. Results: the minimum value of $a_1\cdot a_2\cdot ... \cdot a_{2019}$ occurs for $a_i=a_j, \forall i,j\in\{1,2,\dots,2019\}, i\ne j$, hence for $a_1=a_2=\dots=a_{2019}=b$. $2019\cdot\dfrac{1}{b+2019}=\dfrac{1}{2019}\Longrightarrow b=2019^2-2019=2019\cdot 2018$. $\min{a_1\cdot a_2\cdot ... \cdot a_{2019}}=b^{2019}=2019^{2019}\cdot2018^{2019}$.

Solved with Ryan Yang, Daniel Yuan We claim that $a_1a_2\ldots a_{2019}$ is minimized when they are all equal. We first expand the identity. \begin{align*}&\frac{1}{a_1 + 2019}+\frac{1}{a_2 + 2019}+ ... +\frac{1}{a_{2019} + 2019}=\frac{1}{2019}\\ \iff & 2019\cdot \sum_i \prod_{j\neq i}(a_j+2019) = \prod_i(a_i+2019).\end{align*}Notice that if we move every term which is not $\prod_i a_i$ to the LHS, then everything on the RHS will cancel with something on the LHS since the LHS "overcounts" terms on the RHS. (This is not hard to check.) Thus we may apply AM-GM on each symmetric sum on the LHS. Let the product of all of the $a_i$ be $X$. Suppose that the expanded LHS is at least \[Y = \text{constant}+\sum_{k\in (0,1)}c_k\cdot X^k\]with $c_k$ positive and the constant also positive. Obviously there is a root of \[Y = X\]by IVT since $Y>X$ when $X$ is $0$ and $Y<X$ when we take $X$ arbitrarily large. Consider the smallest root. The inequality is an equality here when all $a_i$ are equal, and this is the minimal $X$ for which the equality can hold for any choice of $a_i$. However, if we fix $a_i=a_j$ for all $i$, $j$, then there is only one solution to the original equation, namely $a_i= 2018\cdot 2019$ for all $i$. Thus we get the answer $X=(2018\cdot 2019)^{2019}$.