Seven positive integers are written in ascending order on the same line. Place five " + " signs and one " = " sign between these numbers to obtain an equality.
Solution 1 - The 7 numbers can be written as

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

Observing that $3 n-6+12=3 n+6$, we conclude 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 and suppose that

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

Since the numbers on the left are smaller, there are more addends on the left, so $k \geq 4$. Assuming $k=4$, the above equality is

$$
4 n+1+2+3+4=3 n+5+6+7
$$

$\log n=8$. In the case $k=5$ we have that

$$
5 n+1+2+3+4+5=2 n+6+7
$$

which does not yield an integer solution. Similarly, $k=6$ does not yield a positive integer solution. Therefore, the only solution is

$$
9+10+11+12=13+14+15
$$