The number 2592009 can be decomposed into the sum of the squares of five positive integers, the sum of these five positive integers is 99, and one of the numbers is 19. Write out this decomposition.
Solution: From $99-19=80$, we know that the sum of the four numbers other than 19 is 80, and the average is 20. Therefore, we can set
$$
\sum_{i=1}^{4}\left(20+a_{i}\right)^{2}+19^{2}=2009,
$$

where, $\sum_{i=1}^{4} a_{i}=0, a_{i}(i=1,2,3,4)$ are integers.
Notice that
$$
400 \times 4+40 \sum_{i=1}^{4} a_{i}+\sum_{i=1}^{4} a_{i}^{2}+19^{2}=2009,
$$

we can see that $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}=48$.
Decomposing 48 into the sum of squares of four integers, we have $48=36+4+4+4$, or other several cases.

For $48=36+4+4+4, a_{i}(i=1,2,3,4)$ possible results are:
(1) One is 6, and the other three are -2;
(2) One is -6, and the other three are 2.
$$
\begin{array}{l}
\text { Then }(20+6)^{2}+(20-2)^{2}+(20-2)^{2}+ \\
(20-2)^{2}+19^{2}=2009,
\end{array}
$$

and $(20-6)^{2}+(20+2)^{2}+(20+2)^{2}+$
$$
(20+2)^{2}+19^{2}=2009 \text {. }
$$

Thus, $2009=26^{2}+19^{2}+18^{2}+18^{2}+18^{2}$ and $2009=22^{2}+22^{2}+22^{2}+19^{2}+14^{2}$ are the required decompositions.

It is easy to see that in other cases, there do not exist $a_{i}(i=1,2,3,4)$ such that $a_{1}+a_{2}+a_{3}+a_{4}=0$, hence no solutions.

In summary, $2009=26^{2}+19^{2}+18^{2}+18^{2}+18^{2}$ and $2009=22^{2}+22^{2}+22^{2}+19^{2}+14^{2}$ are the required decompositions.