310 four-digit number 2011 can be decomposed into the sum of squares of 14 positive integers, among which, 13 numbers form an arithmetic sequence. Write down this decomposition.
Given $1^{2}+3^{2}+\cdots+25^{2}>2011$, we know that the common difference of the arithmetic sequence is less than 2, i.e., the common difference is 1.
Let $2011=y^{2}+\sum_{k=-6}^{6}(x+k)^{2}$.
Then $2011=13 x^{2}+y^{2}+2 \sum_{k=1}^{6} k^{2}$.
Thus, $13 x^{2}+y^{2}=1829$.
Clearly, $x^{2}=\frac{1829-y^{2}}{13}$, so $x$ must be one of the numbers 11, 10, 9, 8, 7.
(1) If $x=11$, then $y^{2}=256 \Rightarrow y=16$. Therefore,
$$
2011=16^{2}+\sum_{k=5}^{17} k^{2}
$$

is one of the required decompositions.
(2) If $x=10$, then $y^{2}=529 \Rightarrow y=23$. Therefore,
$$
2011=23^{2}+\sum_{k=4}^{16} k^{2}
$$

is one of the required decompositions.
(3) When $x=9,8,7$, the corresponding $\dot{y}^{2}$ are 776, 997, 1192, none of which are perfect squares, so they are discarded.
In summary, there are two decompositions that meet the conditions, i.e.,
$$
\begin{array}{l}
2011=16^{2}+\sum_{k=5}^{17} k^{2}, \\
2011=23^{2}+\sum_{k=4}^{16} k^{2} .
\end{array}
$$
(Tian Yonghai, Rao Hua Education Institute, Heilongjiang Province, 152054)