14. (12 points) Xiao Ming wrote the numbers from 1 to 2014 in the following order: first, he wrote 1, then wrote 1 number 2 to the right of 1, and 1 number 3 to the left of 1. Next, he wrote 2 numbers 4, 5 to the right, and 2 numbers 6, 7 to the left. Then, he wrote 3 numbers 8, 9, 10 to the right, and 3 numbers 11, 12, 13 to the left (as shown below):
$\begin{array}{lllllllllllll}\cdots 13 & 12 & 11 & 7 & 6 & 3 & 1 & 2 & 4 & 5 & 8 & 9 & 10 \cdots \text { when writing to } 2014\end{array}$
When writing to 2014, the sum of all the numbers between 1 and 2014 (excluding 1 and 2014) is ( ).
A. 966900
B. 1030820
C. 1989370
D. 2014260
【Solution】Solution: According to the analysis, we find such a pattern, the number of numbers on both sides of 1 is equal, the order of the number of numbers written on the left and right is 1, 2, 3, 4, 5, ...,
After trial calculation, we know: $(1+44) \times 44 \div 2=900$, $990 \times 2=1980$;
When 44 numbers are written on each side of 1, there are a total of 1980 numbers, and 1981 is on the left side of 1.
Next, 45 numbers will be written on each side of 1, and $2014-1980=34$, so from 1982 to 2014 are on the right side of 1.
Then we compare the numbers on both sides of 1,
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline \multicolumn{2}{|c|}{ First Group } & \multicolumn{2}{|c|}{ Second Group } & \multicolumn{2}{|l|}{ Third Group } & \multirow{2}{*}{\begin{tabular}{l\end{array}
} & \multirow{2}{*}{\begin{tabular}{l} 
Forty-fourth \\
Group
\end{tabular}} \\
\hline 3 & 2 & 6,7 & 4,5 & $11,12,13$ & & & \\
\hline \multicolumn{2}{|c|}{ Difference 1} & \multicolumn{2}{|c|}{ Difference $2 \times 2$} & \multicolumn{2}{|l|}{ Difference $3 \times 3$} & $\cdots$ & Difference $44 \times 44$ \\
\hline
\end{tabular}}
\end{tabular}

After writing the 44th group, the sum of all numbers except 1 is: $(2+1981) \times 1980 \div 2=1963170$;
The left side is larger than the right side by: $1 \times 1+2 \times 2+3 \times 3+\cdots+44 \times 44=29370$ (sum of squares formula);
Therefore, when the 44th group is written, the sum on the right side is: $(1963170-29370) \div 2=966900$;
At this point, don't forget to add the sum of 1982 to 2013
When writing to 2014, the total sum on the right side is: $966900+(1982+2013) \times 32 \div 2=966900+63920=1030820$. Therefore, the answer is: $B$.