First, we will take $\pmod{25}.$ Note that by Fermat's Little Theorem, $a^n=a^{n+20}.$ Thus, \begin{align*} 21^{2006}&\equiv21^{6}\pmod{25}\\ &\equiv(-4)^{6}\pmod{25}\\ &\equiv4^6\pmod{25}\\ &\equiv(4^3)^2\pmod{25}\\ &\equiv(64)^2\pmod{25}\\ &\equiv(14)^2\pmod{25}\\ &\equiv196\pmod{25}\\ &\equiv21\pmod{25}. \end{align*}Now, we will take $\pmod{4}.$ Note that \begin{align*} 21^{2006}&\equiv1^{2006}\pmod{4}\\ &\equiv1\pmod{4}. \end{align*}Thus, the last two digits of an integer $x$ that satisfies $x\equiv21\pmod{25}$ and $x\equiv1\pmod{4}.$ From the first congruence, our only possible candidates are $21,46, 71, \text{ and }86.$ It is easy to find that only $21$ satisfies $x\equiv1\pmod{4}.$ We conclude that the last 2 digits of $21^{2006}$ is $\boxed{21}.$