We are distributing some identical candies into three non-empty piles such that the number of candies in each pile is different. How many candies do we have if the number of different possible groupings is exactly one more than the number of candies?

We have a certain number of identical candies, and we need to distribute them into three non-empty piles in such a way that the number of candies in each pile is different. If the number of different possible groupings is exactly one more than the number of candies, how many candies do we have?
The task, formulated in the language of mathematics, means to decompose the natural number $n$ into the sum of 3 different natural numbers. For which $n$ will the number of decompositions be 1 greater than $n$?

Let's write down all possible decompositions for $n=6$:
$\begin{array}{llll}1 & 1 & 4\end{array}$
2
123
$\begin{array}{llll}3 & 1 & 2\end{array}$
$4 \quad 1 \quad 1$
13
$\begin{array}{llll}3 & 2 & 1\end{array}$
$1 \begin{array}{lll}1 & 4 & 1\end{array}$

We can see that some of the listed ones appear more than once, for example, 123 and 213 only differ in the order, which we cannot distinguish because we only have identical types of candies. In the case of $n=6$, there are only three different ways to divide into piles, and only one of them satisfies the condition that each pile has a different number of candies. For $n<6$, none of them can meet the condition of the problem.

Let $n$ be the number of elements ( $n$ pieces of candy), which can be divided into 3 non-empty sets in $\binom{n-1}{2}=\frac{(n-1)(n-2)}{2}$ ways. (Imagine that we place the candies in a row and divide them into three piles with two dividers. The dividers can be placed in $n-1$ gaps.) This is the total number of possible divisions; from this, we need to subtract the number of those that appear more than once or do not meet the requirement of the problem.

Let $k<n$, and let's count the divisions where a set of $k$ ( $k$ pieces of candy) appears twice. These are:

$$
k, \quad k, \quad n-2 k ; \quad k, \quad n-2 k, \quad k ; \quad n-2 k, \quad k, \quad k
$$

This is 3 possibilities, and if $n-2 k=k$, i.e., $n=3 k$ ( $n$ is divisible by 3), then there is one more possibility where each of the 3 piles has the same number of elements (candies).

Accordingly, we need to distinguish 4 cases:

1. If $n$ is even but not divisible by 3, then $n \geq 2 k, k \leq \frac{n}{2}$, and since $k$ is an integer, $\frac{n}{2}-1$ cases exist where two piles have the same number of elements, and each of these appears 3 times. The number of cases that meet our criteria is:

$$
\frac{(n-1)(n-2)}{2}-3\left(\frac{n}{2}-1\right)
$$

Since we do not want to consider the order, we need to divide this by the number of possible orders of 3 elements, $3!=1 \cdot 2 \cdot 3=6$.

Let's write down the equation that satisfies the conditions:

$$
\frac{\frac{(n-1)(n-2)}{2}-3\left(\frac{n}{2}-1\right)}{6}=n+1
$$

Rearranging, we get the equation: $n^{2}-18 n-4=0$, and this has no integer solution.

2. If $n$ is even and also divisible by 3, then

$$
\frac{\frac{(n-1)(n-2)}{2}-3\left(\frac{n}{2}-2\right)-1}{6}=n+1
$$

From this, $n^{2}-18 n=0$, i.e., $n=0$ or $n=18$.

3. If $n$ is odd and not divisible by 3, then $\frac{n-1}{2}$ divisions exist where two piles have the same number of elements. The equation is:

$$
\frac{\frac{(n-1)(n-2)}{2}-3\left(\frac{n-1}{2}\right)}{6}=n+1
$$

$n^{2}-18 n-7=0$. This has no integer solution.

4. If $n$ is odd and divisible by 3, the equation is:

$$
\frac{\frac{(n-1)(n-2)}{2}-3\left(\frac{n-1}{2}-1\right)-1}{6}=n+1
$$

$n^{2}-18 n-3=0$. In this case, there is no integer solution.

We see that $n=18$ is a solution, i.e., 18 pieces of candy can be divided into 3 piles in 19 ways such that the number of candies in each pile is different. There is no other solution to the problem, as we have examined all possible cases.

Based on the work of Gyimesi Andrea (Miskolc, Földes F. Gimn., 8th grade)

Note. Most solvers (most of whom received 3 points) only saw that 18 is a good solution, but not that there is no other solution.