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Thanks for all the very quick responses,they were incredibly useful! Based on these responses, I think the conjecture is now settled in the affirmative, as follows.

For each n, let $F_-(n)$ and $F_+(n)$ be the minimal and maximal values of $\sum_{A \in {\mathcal D}} (-1)^{|A|}$ respectively. The conjecture is that $F_-(n), F_+(n)$ are the extremal values of $(-1)^r \binom{n-1}{r}$ for $r=0,\ldots,n-1$. More explicitly,

$$F_-(n) = -\binom{n-1}{n/2}, F_+(n) = \binom{n-1}{n/2}$$

when $n$ is even,

$$F_-(n) = -\binom{n-1}{(n-1)/2}, F_+(n) = \binom{n-1}{(n+1)/2}$$

when $n=3 \mod 4$, and

$$F_-(n) = -\binom{n-1}{(n+1)/2}, F_+(n) = \binom{n-1}{(n-1)/2}$$

when $n=1 \mod 4$. As mentioned in the post, these bounds would be best possible.

By slicing an n-dimensional downset into two n-1-dimensional downsets, one obtains the inequalities

$$F_-(n-1)-F_+(n-1) \leq F_-(n) \leq F_+(n) \leq F_+(n-1) - F_-(n-1)$$

which already gives most of the conjecture by induction and Pascal's identity; the only remaining cases that need separate verification are

$$F_+(n) = \binom{n-1}{(n+1)/2} \qquad (1)$$

when n is 3 mod 4, and

$$F_-(n) = -\binom{n-1}{(n+1)/2} \qquad (2)$$

when n is 1 mod 4.

Let's show (1), as the proof of (2) is similar. Fix n equal to 3 mod 4, and let ${\mathcal D}$ be a downset which attains the maximal value $F_+(n)$ of $\sum_{A \in {\mathcal D}} (-1)^{|A|}$:

$$\sum_{A \in {\mathcal D}} (-1)^{|A|} = F_+(n).$$

Now introduce the "f-vector" $(f_0,\ldots,f_n)$ of $A$, with $f_i := |\{ A \in {\mathcal D}: |A|=i\}|$ defined as the number of elements of ${\mathcal D}$ of cardinality $i$. (This is shifted by one from the polytope conventions, I guess because i points determine an i-1-dimensional simplex.) Then we have

$$f_0 - f_1 + \ldots - f_n = F_+(n).$$

Let r be the largest index for which $f_r$ is non-zero, or equivalently the largest cardinality of an element of ${\mathcal D}$. (We can treat the degenerate case when ${\mathcal D}$ is empty by hand.) If $r$ was odd, we could simply remove all $r$-element sets from ${\mathcal D}$ and increase the alternating sum, so we may assume that $r$ is even, so the alternating sum looks like $f_0 - f_1 + \ldots - f_{r-1} + f_r$.

The case r=0 can also be treated by hand and will be ignored. Now, we double-count. Observe that each $r$-element set in ${\mathcal D}$ has $r$ "children" as $r-1$-element subsets of ${\mathcal D}$, by removing one of the r elements from that set. On the other hand, each $r-1$-element set can have at most $n-r+1$ "parents", and so

$$r f_r \leq (n-r+1) f_{r+1}.$$f_{r-1}.$$(EDIT: Actually we didn't need to remove the r=0 case if we adopted the convention f_{-1}=0 here.) In particular, if r > \frac{n+1}{2}, then f_r < f_{r-1} we could remove both the r and r-1-element sets from the downset and again increase the sum; so we have r \leq \frac{n+1}{2}. In fact the same argument shows that, by changing the extremum {\mathcal D} if necessary, we may assume that r < \frac{n+1}{2}, thus (since n is 3 mod 4 and r is even) r \leq \frac{n-3}{2}. In other words, every element of {\mathcal D} has cardinality at most (n-3)/2. Now we flip the downset to look at the complementary downset {\mathcal D}' := \{ A \in [n]: [n] \backslash A \not \in {\mathcal D} \}. As n is odd, we have \sum_{A \in {\mathcal D}'} (-1)^{|A|} = \sum_{A \in {\mathcal D}} (-1)^{|A|}, and so {\mathcal D}' is also an extremiser. Thus, by the above argument, every element of {\mathcal D}' has cardinality at most (n+1)/2. Equivalently (as n is odd), {\mathcal D} contains every element of cardinality at most (n-3)/2. Combining this with the previous analysis, we see that the extremum is attained at the set consisting precisely of all subsets of [n] of cardinality at most (n-3)/2, which gives the required value of F_+(n). 1 Thanks for all the very quick responses,they were incredibly useful! Based on these responses, I think the conjecture is now settled in the affirmative, as follows. For each n, let F_-(n) and F_+(n) be the minimal and maximal values of \sum_{A \in {\mathcal D}} (-1)^{|A|} respectively. The conjecture is that F_-(n), F_+(n) are the extremal values of (-1)^r \binom{n-1}{r} for r=0,\ldots,n-1. More explicitly,$$ F_-(n) = -\binom{n-1}{n/2}, F_+(n) = \binom{n-1}{n/2}$$when n is even,$$ F_-(n) = -\binom{n-1}{(n-1)/2}, F_+(n) = \binom{n-1}{(n+1)/2}$$when n=3 \mod 4, and$$ F_-(n) = -\binom{n-1}{(n+1)/2}, F_+(n) = \binom{n-1}{(n-1)/2}$$when n=1 \mod 4. As mentioned in the post, these bounds would be best possible. By slicing an n-dimensional downset into two n-1-dimensional downsets, one obtains the inequalities$$ F_-(n-1)-F_+(n-1) \leq F_-(n) \leq F_+(n) \leq F_+(n-1) - F_-(n-1)$$which already gives most of the conjecture by induction and Pascal's identity; the only remaining cases that need separate verification are$$F_+(n) = \binom{n-1}{(n+1)/2} \qquad (1)$$when n is 3 mod 4, and$$F_-(n) = -\binom{n-1}{(n+1)/2} \qquad (2)$$when n is 1 mod 4. Let's show (1), as the proof of (2) is similar. Fix n equal to 3 mod 4, and let {\mathcal D} be a downset which attains the maximal value F_+(n) of \sum_{A \in {\mathcal D}} (-1)^{|A|}:$$ \sum_{A \in {\mathcal D}} (-1)^{|A|} = F_+(n).$$Now introduce the "f-vector" (f_0,\ldots,f_n) of A, with f_i := |\{ A \in {\mathcal D}: |A|=i\}| defined as the number of elements of {\mathcal D} of cardinality i. (This is shifted by one from the polytope conventions, I guess because i points determine an i-1-dimensional simplex.) Then we have$$ f_0 - f_1 + \ldots - f_n = F_+(n).$$Let r be the largest index for which f_r is non-zero, or equivalently the largest cardinality of an element of {\mathcal D}. (We can treat the degenerate case when {\mathcal D} is empty by hand.) If r was odd, we could simply remove all r-element sets from {\mathcal D} and increase the alternating sum, so we may assume that r is even, so the alternating sum looks like f_0 - f_1 + \ldots - f_{r-1} + f_r. The case r=0 can also be treated by hand and will be ignored. Now, we double-count. Observe that each r-element set in {\mathcal D} has r "children" as r-1-element subsets of {\mathcal D}, by removing one of the r elements from that set. On the other hand, each r-1-element set can have at most n-r+1 "parents", and so$$ r f_r \leq (n-r+1) f_{r+1}.

In particular, if $r > \frac{n+1}{2}$, then $f_r < f_{r-1}$ we could remove both the r and r-1-element sets from the downset and again increase the sum; so we have $r \leq \frac{n+1}{2}$. In fact the same argument shows that, by changing the extremum ${\mathcal D}$ if necessary, we may assume that $r < \frac{n+1}{2}$, thus (since $n$ is 3 mod 4 and r is even) $r \leq \frac{n-3}{2}$. In other words, every element of ${\mathcal D}$ has cardinality at most $(n-3)/2$.

Now we flip the downset to look at the complementary downset ${\mathcal D}' := \{ A \in [n]: [n] \backslash A \not \in {\mathcal D} \}$. As n is odd, we have $\sum_{A \in {\mathcal D}'} (-1)^{|A|} = \sum_{A \in {\mathcal D}} (-1)^{|A|}$, and so ${\mathcal D}'$ is also an extremiser. Thus, by the above argument, every element of ${\mathcal D}'$ has cardinality at most $(n+1)/2$. Equivalently (as $n$ is odd), ${\mathcal D}$ contains every element of cardinality at most $(n-3)/2$. Combining this with the previous analysis, we see that the extremum is attained at the set consisting precisely of all subsets of [n] of cardinality at most $(n-3)/2$, which gives the required value of $F_+(n)$.