2. Determine the smallest natural number that is divisible by 2009 and whose sum of digits equals 2009.
2. Since $2009=223 \cdot 9+2$, the desired number has at least 224 digits. We will consider 224-digit numbers $x=\overline{c_{223} c_{222} \ldots c_{1} c_{0}}$. Clearly, $c_{223} \geq 2$. Moreover, if $c_{223}=2$ then $c_{222}=\cdots=c_{0}=9$ and $x=3 \cdot 10^{223}-1 \equiv 3 \cdot 10-1$, which is not divisible by $2009=7^{2} \cdot 41$ since $x \equiv 1(\bmod 7)$.

Now let $c_{223}=3$. Then the number $x$ has the form $399 \ldots 9899 \ldots 9=4 \cdot 10^{223}-10^{i}-1$ for some $i$. Since $10^{5} \equiv 1(\bmod 41)$, we have $10^{i} \equiv 1,10,18,16$ or 37 $(\bmod 41)$ for $i=0,1,2,3,4(\bmod 5)$ respectively, and thus $x \equiv 4 \cdot 10^{3}-10^{i}-1 \equiv 22-10^{i}$ $(\bmod 41)$ is never divisible by 41.

Let $c_{223}=4$. Among the digits $c_{222}, \ldots, c_{0}$ there are two eights or one seven, while all others are nines; in either case, $x=5 \cdot 10^{223}-10^{i}-10^{j}-1 \equiv 38-\left(10^{i}+10^{j}\right)(\bmod 41)$, where $i$ and $j$ are not necessarily distinct. From the previous, $10^{i}+10^{j} \equiv 38(\bmod 41)$ if and only if $(i, j) \equiv(0,4)$ or $(4,0)(\bmod 5)$. Among other things, $i \neq j$ and $i, j \leq 220$.

Let's try to set $j=220$ and $i \equiv 4(\bmod 5)$. We need to choose $i$, if it exists, such that $7^{2} \mid x=5 \cdot 10^{223}-10^{220}-10^{i}-1 \equiv 5 \cdot 10^{13}-10^{10}-10^{i}-1 \equiv 31-10^{i}$ $(\bmod 49)$. It is easy to find that $10^{i} \equiv 31(\bmod 49)$ if and only if $i \equiv 7$ $(\bmod 42)$, which together with $i \equiv 4(\bmod 5)$ gives the only possibility $i=49$. Therefore, the desired number is

$$
4998 \underbrace{9 \ldots 99}_{170} 8 \underbrace{99 \ldots 9}_{49} .
$$