3.3.

Let the side lengths of $n$ squares be $x_{1}, x_{2}, \cdots, x_{n}$, then $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=2006$.
Since $x_{i}^{2} \equiv 0$ or $1(\bmod 4)$, and $2006 \equiv 2(\bmod 4)$, there must be at least two odd numbers among $x_{i}$.
If $n=2$, then $x_{1}$ and $x_{2}$ are both odd, let them be $2 p+1$ and $2 q+1$, then $(2 p+1)^{2}+(2 q+1)^{2}=2006$.
Thus, $p^{2}+p+q^{2}+q=501$.
However, $p^{2}+p$ and $q^{2}+q$ are both even, which is a contradiction.
If $n=3$, we can set
$$x_{1}=2 p+1, x_{2}=2 k, x_{3}=2 q+1 \text {, }$$

then $(2 p+1)^{2}+(2 k)^{2}+(2 q+1)^{2}=2006$,
which means $p^{2}+p+k^{2}+q^{2}+q=501$.
Clearly, $k$ is odd and $k \leqslant \sqrt{501}$. Hence $k \leqslant 21$.
When $k=1$, $p^{2}+p+q^{2}+q=500$ has no positive integer solutions;
When $k=3$, $p^{2}+p+q^{2}+q=492$ has solutions $p=8, q=20$.
Thus, $17^{2}+6^{2}+41^{2}=2006$. Therefore, $n_{\min }=3$.