Initial 281 Given that there exist $k(k \in \mathbf{N}, k \geqslant 2)$ consecutive positive integers, the mean of their squares is a perfect square. Try to find the minimum value of $k$.


Let the $k$ positive integers be $n+1, n+2, \cdots, n+k (n \in \mathbf{N})$. Then the mean of their squares is
$$
\begin{array}{l}
f(n, k)=\frac{1}{k} \sum_{i=1}^{k}(n+i)^{2} \\
=n^{2}+n(k+1)+\frac{(k+1)(2 k+1)}{6} .
\end{array}
$$

Since $f(n, k) \in \mathbf{Z}$, then $\frac{(k+1)(2 k+1)}{6} \in \mathbf{Z}$.
Thus, $2 \times k, 3 \times k$.
When $2<k \leqslant 13$, we have
$$
\left(n+\frac{k+1}{2}\right)^{2}<f(n, k)<\left(n+\frac{k+1}{2}+1\right)^{2} \text {. }
$$

In fact,
$$
\begin{array}{l}
\text { Equation (1) } \Leftrightarrow n^{2}+n(k+1)+\frac{k^{2}+2 k+1}{4} \\
<n^{2}+n(k+1)+\frac{(k+1)(2 k+1)}{6} \\
<n^{2}+n(k+3)+\frac{k^{2}+6 k+9}{4} \\
\Leftrightarrow \frac{k^{2}+2 k+1}{4}<\frac{2 k^{2}+3 k+1}{6}<\frac{k^{2}+6 k+9}{4}+2 n \\
\Leftrightarrow 3\left(k^{2}+2 k+1\right)<2\left(2 k^{2}+3 k+1\right) \\
<3\left(k^{2}+6 k+9\right)+24 n \\
\Leftrightarrow 1<k^{2}<12 k+25+24 n .
\end{array}
$$

The above inequality is clearly true.
At this point, $\sqrt{f(n, k)} \notin \mathbf{Z}$.
When $14<k<26$, we have
$$
\left(n+\frac{k+1}{2}\right)^{2}<f(n, k)<\left(n+\frac{k+1}{2}+2\right)^{2} \text {. }
$$

In fact,
$$
\begin{array}{l} 
\text { Equation (2) } \Leftrightarrow \frac{k^{2}+2 k+1}{4}<\frac{2 k^{2}+3 k+1}{6} \\
<\frac{k^{2}+10 k+25}{4}+4 n \\
\Leftrightarrow 3\left(k^{2}+2 k+1\right)<2\left(2 k^{2}+3 k+1\right) \\
<3\left(k^{2}+10 k+25\right)+48 n \\
\Leftrightarrow 1<k^{2}<24 k+73+48 n .
\end{array}
$$

The above inequality is clearly true.
At this point, $f(n, k)=\left(n+\frac{k+1}{2}+1\right)^{2}$, i.e., $k^{2}=24 n+12 k+25$.
Then $(k-6)^{2}=24 n+61=8(3 n+7)+5$.
But the remainder of a perfect square when divided by 8 is 0 or 1, which is a contradiction.
When $k=29$,
$$
f(n, 29)=n^{2}+30 n+295=(n+15)^{2}+70 \text {, }
$$

its remainder when divided by 8 is 6 or 7, which is a contradiction.
When $k=31$,
$$
\begin{array}{c}
f(n, 31)=n^{2}+32 n+336 \\
=(n+18)^{2}-4 n+12 .
\end{array}
$$

Taking $n=3$, we get $f(3,31)=21^{2}$.
Therefore, $k_{\text {min }}=31$.
(Zhai Xiaojun, Donghai Senior High School, Jiangsu Province, 222300)