Given 2011 positive integers, the product of which is equal to their sum. How many 1s are there at least among these 2011 numbers?
Let the 2011 positive integers be
$a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{2011}$.
From $a_{1} a_{2} \cdots a_{2011}=a_{1}+a_{2}+\cdots+a_{2011}$
$\leqslant 2011 a_{2011}$,
we know $a_{1} a_{2} \cdots a_{2010} \leqslant 2011a b \Rightarrow(a-1)(b-1)a b \Rightarrow(a-1)(b-1)<1,
\end{array}
$$

Contradiction.
If $A=(2,3,3, a, b)$ or $(3,3,3, a, b)$, the same contradiction is derived.
Therefore, $a_{2004}=1$.
(5) When $a_{2005}=a_{2006}=\cdots=a_{2010}=2, a_{2011}$ $=32$,
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2011}=2004+2 \times 6+32 \\
=2048=a_{1} a_{2} \cdots a_{2011} .
\end{array}
$$

Combining (1)～(5), we know there are at least 2004 ones.
(Liu Donghua, Nankai High School, Tianjin, 300100)