Let $n$ be an even positive integer. A sequence of $n$ real numbers is called complete if for every integer $m$ with $1 \leq m \leq n$, the sum of the first $m$ terms or the sum of the last $m$ terms of the sequence is an integer. Determine the minimum number of integers in a complete sequence of $n$ numbers.
We prove that the minimum number of integers in a complete sequence is equal to 2. First, consider the case $n=2$. Let $a_{1}$ and $a_{2}$ be the numbers in the sequence. Then $a_{1}$ or $a_{2}$ is an integer, say without loss of generality $a_{1}$. Furthermore, $a_{1}+a_{2}$ is an integer, but then $a_{2}$ is also an integer. Therefore, the sequence contains at least two integers.
Now consider the case $n>2$. Write $n=2k$ (since $n$ is even) with $k \geq 2$. Then $a_{1}+a_{2}+\ldots+a_{k}$ or $a_{k+1}+a_{k+2}+\ldots+a_{2k}$ is an integer. Since the sum of both expressions is also an integer, they are both integers. Furthermore, $a_{1}+a_{2}+\ldots+a_{k-1}$ or $a_{k+2}+a_{k+3}+\ldots+a_{2k}$ is an integer. This implies that $a_{k}$ or $a_{k+1}$ is an integer. Additionally, we know that $a_{1}$ or $a_{2k}$ is an integer, and these do not coincide with $a_{k}$ or $a_{k+1}$ because $k \geq 2$. Therefore, at least two different numbers are integers.
Finally, we show that for every even $n$, it is possible to construct a complete sequence with exactly two integers. Write $n=2k$ again. If $k$ is odd, we take $a_{1}=a_{k+1}=1$ and all other terms equal to $\frac{1}{2}$. The sum of all numbers in the sequence is an integer, so it suffices to show that the sum of the first or last $m$ terms is an integer with $1 \leq m \leq k$; the cases with $m>k$ follow directly. For odd $m \leq k$, the first $m$ terms together are an integer, for even $m \leq k$.
We conclude that the minimum number of integers in a complete sequence of $n$ numbers is 2.