Four, (15 points) It is known that a positive integer $n$ can be expressed as the sum of 2011 identical natural numbers, and also as the sum of 2012 identical natural numbers. Determine the minimum value of $n$.


Let $a_{1}, a_{2}, \cdots, a_{2011}$ be 2011 natural numbers with the same digit sum, and $b_{1}, b_{2}, \cdots, b_{2012}$ be 2012 natural numbers with the same digit sum, and they satisfy
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{2011}=n \\
=b_{1}+b_{2}+\cdots+b_{2012} .
\end{array}
$$

Since each of the numbers $a_{1}, a_{2}, \cdots, a_{2011}$ has the same digit sum, each of them has the same remainder when divided by 9, denoted by $r$ $(0 \leqslant r \leqslant 8)$. Similarly, for the numbers $b_{1}, b_{2}, \cdots, b_{2012}$, each has the same digit sum, and thus the same remainder when divided by 9, denoted by $s$ $(0 \leqslant s \leqslant 8)$. Therefore, the numbers $n-2011 r$ and $n-2012 s$ are both multiples of 9. This implies,
$$
\begin{array}{l}
(n-2011 r)-(n-2012 s) \\
=2012 s-2011 r \\
=2012(r+s)-4023 r
\end{array}
$$

is a multiple of 9.
Since 4023 is a multiple of 9 and 2012 is coprime with 9, the number $r+s$ must be a multiple of 9.

If $r=s=0$, then $n \geqslant 9 \times 2012$ (since $b_{1}, b_{2}, \cdots, b_{2012}$ are divisible by 9).

If $r \neq 0$, then $r+s=9$, so at least one of $r \geqslant 5$ or $s \geqslant 5$ holds.
For the number $n$, we get
$$
\begin{array}{l}
n \geqslant 5 \times 2011 \text { and } n \geqslant 5 \times 2012 . \\
\text { Also, } 10055=5 \times 2011 \\
=4 \times 2011+2011 \times 1
\end{array}
$$

and the numbers 4 and 2011 have the same digit sum, so the number 10055 is the desired number.