Not your usual arithmetic sequence problem

First, observe that we can get three overlaps, if the arithmetic sequence includes $4,5,6,7,8,9$ :

$\begin{aligned} \text{4 consecutive terms adding to 30: } &\ldots, 4, 5, \boxed{\textbf{6, 7, 8}, 9}, \ldots \\ \text{5 consecutive terms adding to 30: } &\ldots, \boxed{4, 5, \textbf{6, 7, 8}}, 9, \ldots \end{aligned}$

There also exists one with four overlaps, but we don't actually need it since we want the minimum number of overlaps:

$\begin{aligned} \text{4 consecutive terms adding to 30: } &\ldots, 0, \boxed{\textbf{3, 6, 9, 12}}, \ldots \\ \text{5 consecutive terms adding to 30: } &\ldots, \boxed{0, \textbf{3, 6, 9, 12}}, \ldots \end{aligned}$

We now have to show that it's impossible to have less overlaps. Without loss of generality, suppose that our sequence is increasing, otherwise just reverse it.

Since there are $5$ consecutive terms adding to $30$ , the middle term is $6$ . Thus our sequence has the term $6$ . The sequence cannot be constant, since if it is, everything must be $6$ but then the sum of any $4$ consecutive terms is $24$ , not $30$ . Thus the sequence has difference $d \ge 1$ .

If the overlap is less than $3$ terms, $6$ cannot be included in the $4$ consecutive terms (because it's in the middle of the $5$ consecutive terms; any way you try, using the $6$ requires you to also overlap at least three terms). Thus either all these $4$ consecutive terms are earlier than the $6$ (and thus less than it), or later than the $6$ (and thus greater than it).

The first case is trivial. If all terms are less than $6$ , their sum is less than $6 \cdot 4 = 24$ . This is impossible.

For the second case, observe that the terms are $6+kd, 6+(k+1)d, 6+(k+2)d, 6+(k+3)d$ for some $k \ge 1, d \ge 1$ . Their sum is $24 + (4k+6) d$ . But we also have:

$\begin{aligned} 24 + (4k+6) d &\ge 24 + (4 \cdot 1 + 6) \cdot 1 \\ &= 34 \\ &> 30 \end{aligned}$

Thus this case is also impossible. Thus it's impossible to have less than three overlaps. This gives our answer $\boxed{3}$ .