20. Suppose $x_{1}, x_{2}, \ldots, x_{2011}$ are positive integers satisfying
$$
x_{1}+x_{2}+\cdots+x_{2011}=x_{1} x_{2} \cdots x_{2011}
$$

Find the maximum value of $x_{1}+x_{2}+\cdots+x_{2011}$.
20. Answer: 4022.

Suppose $x_{1}=x_{2}=\cdots=x_{k}=1<2 \leq x_{k+1} \leq \cdots \leq x_{2011}$. Let $M=x_{1} \cdots x_{2011}$. Then
$$
\begin{aligned}
M & =x_{k+1} x_{k+2} \cdots x_{2010} x_{2011} \\
& =\left(x_{k+1}-1\right) x_{k+2} \cdots x_{2010} x_{2011}+x_{k+2} \cdots x_{2010} x_{2011} \\
& \geq\left(x_{k+1}-1\right) 2+x_{k+2} \cdots x_{2010} x_{2011} \\
& \geq \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
& \geq\left(x_{k+1}-1\right) 2+\cdots+\left(x_{2009}-1\right) 2+x_{2010} x_{2011} \\
& \geq\left(x_{k+1}-1\right) 2+\cdots+\left(x_{2009}-1\right) 2+\left(x_{2010}-1\right) 2+\left(x_{2011}-1\right) 2 \\
& =2\left(x_{k+1}+\cdots+x_{2011}-(2011-k)\right) \\
& =2(M-2011) .
\end{aligned}
$$

Therefore, $M \leq 4022$. On the other hand, $(1,1, \ldots, 1,2,2011)$ is a solution to the equation.
So the maximum value is 4022 .