## Task 4 - 281234

Investigate whether there exist 21 pairwise distinct integers and an order $a_{1}, a_{2}, \ldots, a_{21} \quad(*)$

of these numbers such that the following conditions are satisfied:

(1) For any four consecutive numbers in the sequence $\left({ }^{*}\right)$, the sum of these four numbers is negative.

(2) The sum of all 21 numbers is 1989.
We form the sequence $a_{1}, a_{2}, \ldots, a_{21}$, which is recursively defined as follows:

$$
a_{1}=1994, a_{2}=-5989, a_{3}=1995, a_{4}=1996
$$

and for $k=5,6, \ldots, 21$ by

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2249.jpg?height=133&width=579&top_left_y=1024&top_left_x=727)

If we consider the entire sequence, namely

$$
1994,-5989,1995,1996,1997,-5998,1998,1999,2000,-6007,2001,2002,2003,-6016,2004 \text {, }
$$

$$
2005,2006,-6025,2007,2008,2009
$$

and its sum

$$
\begin{gathered}
1994+\underbrace{(-5989)+1995+1996+1997}_{=-1}+\underbrace{(-5998)+1998+1999+2000}_{=-1}+ \\
+\underbrace{(-6007)+2001+2002+2003}_{=-1}+\underbrace{(-6016)+2004+2005+2006}_{=-1}+ \\
+\underbrace{(-6025)+2007+2008+2009}_{=-1}=1994-5=1989
\end{gathered}
$$

the simple construction principle of the sequence should be clear:

By ensuring that the bracketed partial sums are all negative, then according to the construction of the sequence, all other partial sums of any 4 consecutive terms of the sequence are also negative!

Thus, this sequence indeed satisfies all the conditions of the problem.