3. (2007 Croatian Mathematical Competition) Find the maximum natural number $n$ such that $n^{2}+2007 n$ is a perfect square.
3. Let $n^{2}+2007 n=m^{2}\left(m \in \mathbf{N}_{+}\right)$.

Therefore, there exists a positive integer $k$, satisfying $m=n+k$, hence,
$n^{2}+2007 n=(n+k)^{2}$, which means $n=\frac{k^{2}}{2007-2 k}$.
Thus, $2007-2 k>0, k \leqslant 1003$, and
$(2007-2 k) \mid k^{2}$.
To maximize $n$, the numerator $k^{2}$ should be as large as possible, while the denominator should be as small as possible.
Therefore, when $k=1003$, $n=\frac{1003^{2}}{1}=1006009$ is its maximum value.