In how many ways can $1999^{1999}$ be written as the sum of consecutive natural numbers?
Ha $1999^{1999}$ can be written as the sum of consecutive natural numbers, then for some integers $a \geq 0, n \geq 1$,

$$
1999^{1999}=(a+1)+(a+2)+\cdots+(a+n)=n \cdot a+(1+2+\cdots+n)=n \cdot a+\frac{n(n+1)}{2}=\frac{n(n+2 a+1)}{2}
$$

thus $2 \cdot 1999^{1999}=n(n+2 a+1)$.

Since 1999 is a prime number, every divisor of $2 \cdot 1999^{1999}$ is either of the form $1999^{k}(k=0, \ldots, 1999)$ or $2 \cdot 1999^{k}(k=0, \ldots, 1999)$.

If $n=1999^{k}$, then $n+2 a+1>n$ implies $0 \leq k \leq 999$, in which case $n+2 a+1=2 \cdot 1999^{1999-k}$, hence $a=1999^{1999-k}-\frac{1999^{k}+1}{2}$ is indeed positive and an integer, since $1999^{k}$ is odd. In this case, there will be 1000 different $n$ and corresponding $a$ solutions.

If $n=2 \cdot 1999^{k}$, then again $n+2 a+1>n$ implies only $0 \leq k \leq 999$ is possible, in which case $n+2 a+1=1999^{1999-k}$, hence $a=\frac{1999^{1999-k}-1}{2}-1999^{k}$ is again a positive integer, since $1999^{1999-k}$ is odd. Thus, we get 1000 solutions in this case as well.

It is obvious that the 2000 different representations obtained are distinct, and no other representation is possible.