We will show that there are at least $2$ integers in the sequence. Assume the contrary, i.e. assume there are at most $1$ integer.

Let $\{a_i\}_{i=1}^{n}$ be the sequence.

For $m=1$, we get either $a_1$ or $a_n$ is integer. WLOG, $a_1$ is integer.

For $m=2$, either $a_1+a_2$ or $a_{n-1}+a_n$ is integer. If $a_1+a_2$ is integer, then $a_2$ must be an integer which contradicts our assumption. So, $a_{n-1}+a_n \in \mathbb{Z}$

Similarly for $m=3$, we get $a_2+a_3 \in \mathbb{Z} $

Analogously, we get some pairwise sums which are integers.

For $m=n$, $a_1 + a_2+ a_3 +\cdots + a_{n-1} + a_n \in \mathbb{Z} \Rightarrow a_2+ a_3 +\cdots + a_{n-1} + a_n \in \mathbb{Z}$

$n-1$ is odd. So we can't divide the numbers $a_2, a_3, \cdots, a_{n-1}, a_n$ into some pairs.

So, $a_1 + a_2+ a_3 +\cdots + a_{n-1} + a_n$ can't be an integer if there are at most $1$ integer.

Contradiction!

Now, all we need to do is to find such a construction where there are $2$ integers and all our conditions are satisfied.

When, $n=4k$ our construction will be \[\underbrace {1, .5, .5, \cdots .5, 1}_{first 2k terms} , \underbrace {.5 .5, .5, \cdots .5, .5}_{second 2k terms} \]

When, $n=4k+2$ our construction will be \[\underbrace {1, .5, .5, \cdots .5, .5}_{first 2k terms} , \underbrace {1 .5, .5, \cdots .5, .5}_{second 2k terms} \]

So, the answer is $2$