Solution: Let $n$ be any positive integer.
When $n=2 k+1\left(k \in \mathbf{N}_{+}\right)$, we have $n=2 k+1=(k+1)^{2}-k^{2}$.
Thus, every odd number greater than 1 is a wise number.
When $n=4 k\left(k \in \mathbf{N}_{+}, k \geqslant 2\right)$, we have
$$n=4 k=(k+1)^{2}-(k-1)^{2} \text {. }$$

Thus, every number greater than 4 and divisible by 4 is a wise number.
When $n=4 k-2\left(k \in \mathbf{N}_{+}\right)$, let $n=p^{2}-q^{2}$.
Then $2(2 k-1)=(p+q)(p-q)$.
Thus, $(p+q)(p-q)$ is even.
Since $p+q$ and $p-q$ have the same parity, both $p+q$ and $p-q$ are even. Therefore, $4 \mid (p+q)(p-q)$. Hence, $4 \mid 2(2 k-1)$, which implies $2 \mid (2 k-1)$, a contradiction.

Therefore, every number of the form $4 k-2\left(k \in \mathbf{N}_{+}\right)$ is not a wise number.

If all wise numbers are arranged in ascending order and, except for the first wise number, the rest are grouped into sets of 3:
$$3,(5,7,8),(9,11,12), \cdots$$

According to the above discussion, the last number in each set is a multiple of 4, so the $k$-th set is
$$(4 k+1,4 k+3,4 k+4) \text {. }$$

Since $2009-1=3 \times 669+1$, the 2009th position in the sequence is the first number of the 670th set, which is $4 \times 670+1=2681$.