If we add up the natural numbers from 1 to $n$, the sum in certain cases can be divisible by 10 raised to some positive integer power. What is the smallest $n$ for which the sum is divisible by ten thousand?
We know that the sum of the first $n$ natural numbers is $S_{n}=\frac{n(n+1)}{2}$. According to the problem, we need to find the smallest natural number $n$ such that

$$
\frac{n(n+1)}{2}=k \cdot 10^{4}
$$

and rearranged:

$$
n(n+1)=k \cdot 2 \cdot 2^{4} \cdot 5^{4}=k \cdot 2^{5} \cdot 5^{4}=k \cdot 32 \cdot 625
$$

with an appropriate natural number $k$.

$n$ and $n+1$ are two consecutive integers, so they cannot have a common divisor greater than 1. Therefore, only one of them can be even; and that one must be divisible by 32. Similarly, only one of them can be divisible by 5; which means it must also be divisible by 625.

It is possible that the same number is divisible by both 32 and 625, which gives $n=32 \cdot 625=20000$, or $n+1=32 \cdot 625$, so $n=19999$. However, it is also possible that one of $n$ and $n+1$ is divisible by 32 and the other by 625. This represents two cases, depending on whether $n$ or $n+1$ is divisible by 32.

Let's first consider $n=32 x$ and $n+1=625 y$, with suitable integers $x$ and $y$. Thus, $625 y=32 x+1$, or in standard form:

$$
625 y-32 x=1
$$

This is a so-called two-variable linear Diophantine equation. Its general form is $a \cdot x+b \cdot y=c$, where $a, b, c$ are given integers; and the equation is Diophantine because we seek solutions $(x, y)$ in the integers. It is known that such equations have solutions if and only if the greatest common divisor of $a$ and $b$ divides $c$. In this case, this is clearly satisfied. Now we will solve the equation in this specific case. We consider the smaller of the coefficients of the unknowns, which is 32, and find the number closest to 625 that is divisible by 32, which is 640. Writing the initial equation as $640 y-15 y-32 x=1$ and factoring out the 32, we get $32(20 y-x)-15=1$, and we introduce the new unknown $u=20 y-x$, where $u$ is also an integer; thus we have:

$$
32 u-15 y=1
$$

This is a Diophantine equation for $u$ and $y$. Using a similar procedure as before, we write this as $15(2 u-y)+2 u=1$, which gives:

$$
15 v+2 u=1, \quad \text{where} \quad v=2 u-y \text{ is an integer.}
$$

We need one more step: we write the equation as $2(7 v+u)+v=1$, or:

$$
2 w+v=1, \quad \text{where} \quad w=7 v+u \text{ is an integer.}
$$

Now we can express $v$ in terms of $w$: $v=1-2 w$. Then, using the relations (3), (2), and (1), we can express $u, y$, and $x$ in terms of $w$:

$u=w-7 v=w-7(1-2 w)=15 w-7, y=2 u-v=2(15 w-7)-(1-2 w)=32 w-15, x=20 y-u=20(32 w-15)-(15 w-7)$

From the procedure, it follows that $x$ and $y$ can only be of this form, where $w$ is an integer. Conversely, if $w$ is an integer, then substituting the values of $x$ and $y$ into the initial equation, they satisfy it. Since $n$ is positive, $x$ and $y$ must also be positive, which is only possible if $w$ is positive. The smallest $n$ is obtained when $x$ (and $y$) are minimal. This happens when $w=1$, so $x=332$ and $y=17$. This gives $n=332 \cdot 32=10624$. (Similarly, $n+1=17 \cdot 625=10625$).

We must not forget the other case, where $n=625 y$ and $n+1=32 x$. This leads to the equation $625 y-32 x=-1$. Performing the procedure here, we get:

$$
y=32 w+15 \quad \text{and} \quad x=625 w+293
$$

Now, for $n$ (and thus $x$ and $y$) to be positive, it is sufficient that $w \geq 0$; and clearly the minimal possibility is when $w=0$. We get:

$$
n=625 \cdot y=625 \cdot 15=9375
$$

which indeed provides the solution, since $9375<10624<19999<20000$.

Remarks. 1. Those who found the solution by trial and error, provided they found all the correct values, also received the maximum 5 points.

2. Those who stated that the smallest value of $n$ is 20000 did not receive any points for their solutions.