320 Suppose there are $2 k+1$ consecutive natural numbers,
where the sum of the squares of the $k$ larger natural numbers is equal to the sum of the squares of the remaining $k+1$ smaller natural numbers (such as $5^{2}=4^{2}+3^{2}$ $\left.(k=1), 14^{2}+13^{2}=12^{2}+11^{2}+10^{2}(k=2)\right)$. For convenience, such an equation is called a "modified way". Clearly, the natural numbers $3, 4, 5, 10, 11, 12, 13, 14$ are all numbers in the modified way. The question is: Is the four-digit number 2012 a number in the modified way?
Let
\[
\begin{array}{l}
(a+k)^{2}+(a+k-1)^{2}+\cdots+(a+1)^{2} \\
=a^{2}+(a-1)^{2}+\cdots+(a-k)^{2}
\end{array}
\]

be the $k$-th transformation method,
\[
w_{k}=(a+k, a-k)
\]

indicates that the largest number is $a+k$ and the smallest number is $a-k$.
Clearly, $a^{2}=4 \times(1+2+\cdots+k) a$, or $a=4 \times(1+2+\cdots+k)=2 k(k+1)$.

From $2 k(k+1)-k \leqslant 2012 \leqslant 2 k(k+1)+k$, solving gives $k=31$.
It is easy to see that $w_{31}=(2015,1953)$.
Thus, 2012 is the 31st transformation method number.
(Tian Yonghai, Suihua Educational Institute, Heilongjiang, 152054)