Ten consecutive positive integers have a sum that is a divisor of the sum of their squares. Which are these ten numbers?
Let the 10 consecutive positive integers be $n, n+1, n+2, \ldots, n+9$. Then the sum of these ten numbers is: $n+(n+1)+\cdots+(n+9)=$ $10 n+45$, and the sum of their squares is:

\[
\begin{gathered}
n^{2}+(n+1)^{2}+\cdots+(n+9)^{2}= \\
=10 n^{2}+2 n+4 n+6 n+\cdots+18 n+1^{2}+2^{2}+3^{2}+\cdots+9^{2}=10 n^{2}+90 n+285
\end{gathered}
\]

Thus, the value of the fraction $\frac{10 n^{2}+90 n+285}{10 n+45}$ must be an integer.

\[
\frac{10 n^{2}+90 n+285}{10 n+45}=n+4+\frac{n+21}{2 n+9}
\]

Therefore, $\frac{n+21}{2 n+9}$ must be an integer.

Since $n$ is a positive integer, $n+21<6 n+27$, so $\frac{n+21}{2 n+9}<3$, which means $\frac{n+21}{2 n+9}=1$ or $\frac{n+21}{2 n+9}=2$ are the only possibilities. In the first case, $n=12$, and in the second case, $n=1$, both of which are positive integers.

Thus, the 10 consecutive positive integers are either $1,2,3, \ldots, 10$ or $12,13,14, \ldots, 21$.