Seven positive and consecutive integers are written in ascending order on the same line. Determine if it is possible to place five "+" signs and only one "=" sign between these numbers in such a way that the result is an equality.
Solution 1: The seven numbers can be written as

$$
\underbrace{n-3, n-2, n-1}_{3 n-6}, n, \underbrace{n+1, n+2, n+3}_{3 n+6}
$$

Noting that $3 n-6+12=3 n+6$, we establish that $n=12$. Therefore, the numbers are $9+10+11+12=13+14+15$.

Solution 2: Let $n+1, n+2, \ldots, n+7$ be the seven consecutive numbers such that

$$
(n+1)+\cdots+(n+k)=(n+k+1)+\cdots+(n+7)
$$

for some $k$ between 1 and 6. Since all the numbers on the left side of the equation are smaller than the numbers on the right, there must be more terms on the left, so $k=6, 5$ or 4. Taking $k=6$, we get $6 n+1+2+3+4+5+6=n+7$, that is, $5 n+14=0$, which has no integer solution. Also, $k=5$ gives $5 n+1+2+3+4+5=n+6+n+7$, that is, $3 n+2=0$, which has no integer solution. Finally, with $k=4$ we get $4 n+1+2+3+4=3 n+5+6+7$, so $n=8$ and $9+10+11+12=13+14+15$ is the only solution.