In this sentence, the number of numbers occurring 1 time is $a_{1}$, the number of numbers occurring 2 times is $a_{2}$, $\ldots$, the number of numbers occurring 2013 times is $a_{2013}$. Determine the numbers $a_{1}, a_{2}, \ldots, a_{2013}$ so that we get a true statement. In how many ways can we do this?
Solution. For easier formulation, let's introduce the notation $b_{i}=i$ ( $1 \leqslant i \leqslant 2013$ ), so the problem can be restated as follows: the number of numbers that occur $b_{1}$ times is $a_{1}$, the number of numbers that occur $b_{2}$ times is $a_{2}$, and so on. If $a_{n}=1$, then - since all $b_{i}$ are distinct - there is a number $k$ that appears $n$ times in the statement (i.e., among the $a_{i}$ and $b_{i}$) in total, so there are exactly $n-1$ instances of $k$ among the $a_{i}$ - except if $k=0$, in which case there are exactly $n$, since there is no zero among the $b_{i}$. If $a_{\ell}=m$, where $m$ is a positive integer, then for the statement to be true, there must exist $m$ distinct numbers, each of which appears at least $\ell-1$ times among the $a_{i}$. We also know that the sum of the products of the corresponding $b_{i}$ and $a_{i}$ numbers is exactly 4026, since a total of this many numbers appear in the statement. Let's break down the problem into cases based on how many zeros there are among the $a_{i}$ numbers.

1. Case: Exactly 2008 zeros. Then the value of $a_{2008}$ is at least 1. In addition to this, there are four other $a_{i}$ numbers that are not zero, let these be $a_{c}, a_{d}, a_{e}$, and $a_{f}$, with values $p, q, r$, and $s$ - where $p, q, r, s \geq 1$. Thus, there are at least $p(c-1)+q(d-1)+r(e-1)+s(f-1)$ numbers among the $a_{i}$ that are not zero. This is at least $1 \cdot(1-1)+1 \cdot(2-1)+1 \cdot(3-1)+1 \cdot(4-1)=6$ numbers, since $c, d, e, f$ are distinct. We have reached a contradiction, so there can be no such solution.
2. Case: Fewer than 2008 zeros. Then we can add a sixth non-zero $a_{j}$ number to the five non-zero $a_{i}$ numbers from the previous case - let this be $a_{g}$; thus, the largest of the five smallest $i$ for which the $a_{i}$ are non-zero is at least 5, since they are distinct. Therefore, the number of non-zero $a_{i}$ numbers has increased by more than 1, at least $(5-1) \cdot 1$. Following this line of thought, we find that if we decrease the number of zeros one by one, the number of non-zero $a_{i}$ numbers increases by more than one at a time, so there is no solution to the problem in this case either.
3. Case: Exactly 2013 zeros. This is obviously not possible, since it would mean that every $a_{i}$ is zero, i.e., there is no number that appears $1,2, \ldots, 2013$ times; in this case, every number in the statement would have to appear at least 2014 times. However, this would not be true for zero.
4. Case: Exactly 2012 zeros. Then the only $a_{i} \neq 0$ value is $a_{2012}=k \geq 1$, so the total of $k$ different numbers each would have to appear exactly 2012 times, hence $2012 \cdot k=4026$, which is a contradiction.
5. Case: Exactly 2011 zeros. Then $1 \leq a_{2011}=k \leq 2$, and there is only one index $i \neq 2011$ for which $a_{i} \neq 0$; this is clearly $a_{1}$, and $a_{1} \geq 2010$. Thus, $4026=2011 \cdot k+1 \cdot a_{1}$, so $k=1$. However, this means that 1 appears exactly twice, so $a_{2} \neq 0$, which is a contradiction.
6. Case: Exactly 2010 zeros. Similarly to the previous case, $a_{2010}=1$ and $a_{1} \neq 0$. There cannot be any other $a_{t}$, $t \geq 5$ number that is not 0, since then there would be at least $5-1$ $a_{i}$ that are not 0. If neither $a_{3}$ nor $a_{4}$ is 0, or one is 0 and the other is more than 1, then there are at least $2 \cdot(3-1)$ $a_{i}$ that are not 0. If $a_{4}=1$, then $a_{1}=1$, which is not possible, since there are at least 2010 numbers that appear exactly once, since there are 2010 equal $a_{i}$. If $a_{3}=1$, then $a_{1}=u$, where $2013 \geq u \geq 2010$, so $u$ appears exactly twice, so $a_{2}$ is not 0. Here, only the case remains where $a_{1}$, $a_{2}$, and $a_{2010}$ are not zero, then $1 \cdot a_{1}+2 \cdot a_{2}+a_{2010} \cdot 1=4026$. So $a_{1} \cdot 1+a_{2} \cdot 2=2016$. Here, $a_{2}=1$ is not possible, in the case of $a_{1}=2014, a_{2}=2$ then $a_{1}=2012$, which is not possible, since there are more than 1 of both 2 and 1. However, $a_{2}=3$ and $a_{1}=2010$ is a good solution. In this case, 0 appears 2010 times, 2010, 1, and 3 appear twice, and the other (2010) numbers appear once.
7. Case: Exactly 2009 zeros. This time $a_{2009}=1$. Following the previous reasoning, the second largest $a_{i}$ cannot be more than 5. It cannot be exactly five either, because then - since $a_{1} \geq 2009$ - the sum of the products $a_{i} b_{i}$ would be at least

$$
1 \cdot 2009+2 \cdot 1+3 \cdot 1+5 \cdot 1+2009 \cdot 1
$$

which is too much. If the second largest $a_{i}=4$, then the product sum is at least $1 \cdot 2009+2 \cdot 1+4 \cdot 1+2009 \cdot 1=4026$, which is just right, and this minimum can only be achieved if $a_{1}=2009, a_{2}=1, a_{4}=1$ and $a_{2009}=1$. Then 1 appears four times, 2009 appears twice, 0 appears 2009 times, and the other numbers appear once, which is a good solution. If the second largest $a_{i}$ is 3, then the product sum can be made true with $a_{1}=2009, a_{2}=1, a_{3}=2$ and $a_{2009}=1$, or $a_{1}=2010$, $a_{2}=2, a_{3}=1$ and $a_{2009}=1$. Only the latter is a valid solution. If $a_{1}=2011$, then $a_{2}=1$, $a_{3}=1$, which is a contradiction, since there are too many 1s. Clearly, $a_{1}$ cannot be more than 2011 either.

With this, we have reviewed all cases, so the statement can be made true in three ways.