Initially 247 Suppose there are $n$ square cards of the same size, each inscribed with a positive integer. These $n$ positive integers are consecutive two-digit numbers, and the sum of these two-digit numbers equals the "composite number" of the first and last terms (such as $1+2+\cdots+5=15,4+5+\cdots$ $+29=429$). If the side length of the cards is an integer greater than 1, and the sum of the areas of these cards is 2009, then, what numbers are inscribed on the cards?
Solution: Clearly, $2009=41 \times 7^{2}$.
Given that the side lengths of the cards are integers greater than 1, and the sum of the areas of these cards is 2009, it follows that the area of each card is 49, and the total number of cards is $n=41$.
Let the numbers on the cards be
$$
\begin{array}{l}
a_{1}=10 a+b, a_{2}=10 a+b+1, \\
a_{3}=10 a+b+2, \cdots \cdots \\
a_{41}=10 a+b+40 .
\end{array}
$$

Given that the sum of these positive integers equals the number formed by the first and last terms, we have
$$
\begin{array}{l}
\frac{41(10 a+b+10 a+b+40)}{2} \\
=1000 a+100 b+10 a+b+40 .
\end{array}
$$

Simplifying, we get
$$
60 a+6 b=78 \text { or } 10 a+b=13 \text {. }
$$

Clearly, $a_{1}=10 a+b=13$,
$$
\begin{array}{l}
a_{2}=10 a+b+1=14, \\
a_{3}=10 a+b+2=15, \\
\cdots \cdots .
\end{array}
$$
$$
a_{41}=10 a+b+40=53 \text {. }
$$

Noting that $13+14+\cdots+53=1353$, the numbers on the cards are: $13,14, \cdots, 52,53$.