Three, (25 points) On the first day of operation, the factory's production does not exceed 20 units, and the daily production increases thereafter, but the increase in production each time does not exceed 20 units. When the daily production reaches 1996 units, find the minimum value of the total production of the factory.

When the daily production reaches 1996 units, find the minimum value of the total production of the factory.
Three, suppose the production on the $n$-th day after the start of work reaches 1996 pieces, and the production on the first day is $a_{1}$ pieces, with the increase in production on the $i$-th day being $a_{i}$ pieces $(i=2,3, \cdots, n)$, then we have
$$
a_{1}+a_{2}+a_{3}+\cdots+a_{n}=1996,0<a_{i} \leqslant 20 .
$$

The total production over these $n$ days is
$$
S=n a_{1}+(n-1) a_{2}+(n-2) a_{3}+\cdots+a_{n} \text {. }
$$

The problem is to find the minimum value of (2) under condition (1).
In $a_{1}, a_{2}, \cdots, a_{k}, \cdots, a_{n}$, counting from the end, let the first number less than 20 be $a_{k}(1<k \leqslant 20)$, at this time, make the following adjustments to these $n$ numbers, let
$$
\begin{array}{l} 
a_{1}^{\prime}=a_{1}-1, a_{2}^{\prime}=a_{2}, a_{3}^{\prime}=a_{3}, \cdots, a_{k-1}^{\prime}=a_{k-1}, a_{k}^{\prime}=a_{k} \\
+1, a_{k+1}^{\prime}=a_{k+2}^{\prime}=\cdots=a_{n}^{\prime}=20 .
\end{array}
$$

We still have $a_{1}^{\prime}+a_{2}^{\prime}+a_{3}^{\prime}+\cdots+a_{n}^{\prime}=1996$.
However, the corresponding $S^{\prime}$ is smaller than the original $S$, i.e.,
$$
S-S^{\prime}=k-1 \text {. }
$$

It can be seen that the minimum value of $S$ is obtained when $0<a_{1} \leqslant 20$, and $a_{2}$
$$
\begin{array}{l}
=a_{3}=\cdots=a_{n}=20 \text {, since } \\
1996=99 \times 20+16 \text {, } \\
\text { hence when } a_{1}=16, n=100 \text {, } \\
S_{\text {min }}=100 \times 16+(1+2+\cdots+99) \times 20 \\
=100600 \text {. } \\
\end{array}
$$