Define the profile of a stable set to be the set of points in it with even $x$-coordinate. It suffices to show that among all subsets of $S$ with the same profile, at least half have even cardinality. [asy][asy] size(100); int[] s = {11, 8, 8, 6, 4, 4, 3, 2, 2, 1}; int[] p = {0, 7, 0, 5, 0, 2, 0, 2, 0, 0}; for (int i = 1; i <= s.length; ++i) { for (int j = 1; j <= s[i-1]; ++j) { if (j <= p[i-1]) { dot((i, j), red); } else { dot((i, j)); if (i % 2 == 1 && j <= (i > 1 ? p[i-2] : s[0]) && j > (i < s.length ? p[i] : 0)) { draw(shift((i, j)) * ((-0.5, -0.5) -- (-0.5, 0.5) -- (0.5, 0.5) -- (0.5, -0.5) -- cycle)); } } if (j == (i > 1 ? min(s[i-1], p[i-2]) : s[0])) { label(scale(0.5) * ("$c_" + string(i) + "$ = " + string((i > 1 ? p[i-2] : s[0]) - (i < s.length ? p[i] : 0))), (i+1, s[i-1]+1.25)); } if (i % 2 == 0 && p[i-1] != 0) { draw((i - 1.5, p[i-1]+0.5) -- (i + 0.5, p[i-1]+0.5) -- (i + 0.5, (i < s.length ? p[i+1] : 0)+0.5)); } } } label("$P$", (1.5, -0.5), red); label("$S$", (-1, s[0]+0.5)); [/asy][/asy] Fix a profile $P$, and for each odd-indexed column $i$, let $c_i$ be the number of points in the column with a point in $P$ (or no point) to its immediate left but not to its right. Then each stable subset with profile $P$ can be represented as an element of \[\{0, 1, \ldots, c_1\} \times \{0, 1, \ldots, c_3\} \times \{0, 1, \ldots, c_5\} \times \ldots,\]and the subset has even size if and only if its corresponding tuple has even sum. If a $c_i$ is odd, then exactly half of the tuples have even sum. If every $c_i$ is even, then there is exactly one more tuple with even sum than with odd sum, by induction. $\blacksquare$

Definitely not as elegant as the above solution. Call the border of a stable set $S$ the set $$\{(x,y) \in S \ | \ (x+1,y+1) \not\in S \}.$$ We induct on the number of points. If there is either $0$ or $1$ points, then the conclusion is clear, so the base case has been established. Also assume if $S$ is stable and $|S| < n$ then the conclusion follows. For the inductive step assume $|S|=n$. Choose a point $P=(x_0,y_0)$ in the border of $S$ with at least one even coordinate, which clearly exists if $n > 1$. (We can just take a point with $x$-coordinate or $y$-coordinate equal to $2$.) We will do cases on whether $P$ is in the stable subset or not. Case 1: $P$ is in the subset. Then all points $(x,y)$ with $x<x_0,y<y_0$ are in the subset. There are $x_0y_0$ of these, which is even. The remaining points in $S$ can be split into two (possibly empty) sets, $S_1$ and $S_2$. [asy][asy] fill((0,0)--(0,7)--(4,7)--(4,0)--cycle, green + opacity(0.02)); fill((4,5)--(6,5)--(6,2)--(8,2)--(8,0)--(4,0)--cycle,cyan + opacity(0.02)); fill((0,7)--(0,10)--(3,10)--(3,8)--(4,8)--(4,7)--cycle,cyan+opacity(0.02)); draw((0,10)--(3,10)--(3,8)--(4,8)--(4,5)--(6,5)--(6,2)--(8,2)--(8,0),red); draw((8,0)--(0,0)--(0,10)); dot("$P$",(3.5,6.5),SW); label("$S_1$",(5,1)); label("$S_2$",(1,8)); [/asy][/asy] Note that both $S_1$ and $S_2$ are stable if they were to be translated accordingly. Let $S_i$ have $E_i$ "stable" subsets of even size, and $O_i$ of odd size. Since we have already included an even number of points in our subset, we get $E_1E_2+O_1O_2$ stable subsets of even size, and $E_1O_2+E_2O_1$ stable subsets of odd size. As $$(E_1E_2+O_1O_2)-(E_1O_2+E_2O_1)=(E_1-O_1)(E_2-O_2) \geq 0$$by the inductive hypothesis, this case gives at least half of stable subsets are of even size. (The last step is also just the rearrangement inequality in $2$ variables.) Case 2: $P$ is not in the subset. The points directly above $P$ as well as those directly to the right of $P$ are then also not in the subset. We may delete these points, and we are left with another stable set $S'$, and at least half of stable subsets of $S'$ are of even size by inductive hypothesis. Thus, both cases give at least half of stable subsets having even size, so overall at least half of stable subsets are of even size.

Please someone check if my solution is correct or not...

@below,Yes you re right.Thanks for pointing out the mistake

@above unfortunately, your mapping from odds to evens is not injective, I believe the diagram below shows that a 3x3 pillar followed by a 2x2 pillar and a 3x5 pillar would both map to a 2x5 pillar.

Poorly written, needs a diagram Consider every stable set, and for each $(x, y)$ in the set colour in the square $(x, y)$ on an infinite array, where the bottom left square is $(0, 0)$. Then, by the condition, the stable set is all the points underneath some path from the y axis to the x axis, such that every step moves one step to the right or one step down. Assume the first move is rightward and the last move is downward. Denote this as the "path of S". We will now prove the result via induction on the number of elements in $S$. Our base case is if $S$ had $0$ elements, so it is empty. Then, since the empty set has an even number of elements, at least half of all the stable subsets of $S$ has an even number of elements, so our base case is proven. Now, assume for all stable sets $|T| < |S|$, at least half the stable subsets of $T$ has an even number of elements. Clearly any stable subset of $S$ must have it's path be below and to the left of the path of $S$. Consider a point $P = (x, y)$ on the path such that $xy$ is even. We will further impose the condition that $P$ is at most $2$ steps away from the end of the path, so either we move right then down to get to the end, down and down, or just down. One of these points must have at least one even coordinate (and be on the path). Then, for any stable subset $T$ of $S$ whose path does not contain $P$, we can create a new path by removing every square whose coordinates $(x', y')$ are $x'\geq x, y'\geq y$, then adjusting the new path to go around these removed squares borders, which will also have $T$ as a stable subset. By our inductive hypothesis, at least half of these paths have even elements. Then, if a path contained $P$, then it must also contain the rectangle with corners at $(0, 0), (x, 0), (x, y), (0, y)$, which has an even number of elements. Furthermore, the beginning of the path will go from the y axis to $P$, so if cut the path off by the line parallel to the $x$ axis and through $P$, by our inductive hypothesis at least half of these sub-paths will contain an even number of elements. Finally, the last bit of the path, from $P$ to the end, either all of them contain even elements or exactly half of them do (again, via our condition of $P$). Thus, for the paths that pass through $P$, at least half of them contain even elements. Since in both cases at least half the paths have even elements, we conclude that over all stable subsets, at least half of them have even elements. Our induction is complete.

Suppose $S$ has maximum $x$-coordinate equal to $n$. Note that any stable subset of $S$ can be seen as a sequence of integers $0 \le a_1 \le a_2 \le \cdots \le a_n$ where $a_i \le b_i$ (where the set is represented by a nondecreasing sequence $b_1,b_2,\cdots,b_n$ by looking at the $y$-coordinates from right to left). The number of elements in the subset is just $\sum a_i$. We use dynamic programming. Let $E(i,j)$ denote the number of nondecreasing sequences $c$ of length $i$ that ends with $j$ such that $c_k \le b_k$ for all $k \le i$ and $\sum c_k$ is even. Define $O(i,j)$ similarly (but for when $\sum c_k$ is odd). Let $dp(i,j)=E(i,j)-O(i,j)$. We have $dp(0,0)=1$ and $dp(0,x)=0$ for all other $x$. By considering all possible values of the last number in the sequence, we can obtain the recurrence \[dp(i,j) = (-1)^{j}\sum_{k=0}^{j}dp(i-1,k)\]for all $j \le b_i$. Define $dp(i,j)=0$ elsewhere. The key claim is that for all $i,j$, we have $dp(i,2j) \ge 0$, $dp(i,2j+1) \le 0$ and $|dp(i,2j+1)| \le |dp(i,2j)|$. We will prove this by induction. Observe that this is true for $i=0$. Suppose it is true for all smaller $i$. If $2j > b_i$, we are done. Otherwise, letting $S=\displaystyle\sum_{k=0}^{2j-1}dp(i-1,k)$, $a = dp(i-1,2j) \ge 0$ and $b = -dp(i-1,2j+1) \ge 0$ (with $a \ge b$), we obtain $dp(i,2j) = S+a$, $dp(i,2j+1)=-(S+a-b)$ (or $0$ if $2j+1>b_i$). In any case, $dp(i,2j) \ge 0$, $dp(i,2j+1) \le 0$ and $|dp(i,2j+1)| \le |dp(i,2j)|$, as desired. It remains to note that the difference between the number of stable subsets of $S$ with even elements and those with odd elements is just $\displaystyle\sum_{k=0}^{b_n}dp(n,k)$, which is clearly nonnegative by our claim.

Current draft of official solution, given by Nikolai Beluhov: We proceed by induction on $|S|$, with $|S| \le 1$ clear. Suppose $|S| \ge 2$. For any $p \in S$, let $R(p)$ denote the stable rectangle with upper-right corner $p$. We say such $p$ is pivotal if $p + (1, 1) \notin S$ and $|R(p)|$ is even. [asy][asy] path R = box( (0,0), (6,3) ); fill(R, palered); fill(shift(5,2)*unitsquare, lightcyan); label("$p$", (5.5,2.5)); for (int i=0; i<=0; ++i) { draw(shift(i,5)*unitsquare); } for (int i=0; i<=1; ++i) { draw(shift(i,4)*unitsquare); } for (int i=0; i<=3; ++i) { draw(shift(i,3)*unitsquare); } for (int i=0; i<=7; ++i) { draw(shift(i,2)*unitsquare); } for (int i=0; i<=10; ++i) { draw(shift(i,1)*unitsquare); } for (int i=0; i<=10; ++i) { draw(shift(i,0)*unitsquare); } draw(R, heavyred+1.5); [/asy][/asy] Claim: If $|S| \ge 2$, then a pivotal $p$ always exists. Proof. Consider the top row of $S$. If it has length at least $2$, one of the two rightmost points in it is pivotal. Otherwise, the top row has length $1$. Now either the top point or the point below it (which exists as $|S| \ge 2$) is pivotal. \qedhere $\blacksquare$ We describe how to complete the induction, given some pivotal $p \in S$. There is a partition \[ S = R(p) \sqcup S_1 \sqcup S_2 \]where $S_1$ and $S_2$ are the sets of points in $S$ above and to the right of $p$ (possibly empty). Claim: The desired inequality holds for stable subsets containing $p$. Proof. Let $E_1$ denote the number of even stable subsets of $S_1$; denote $E_2$, $O_1$, $O_2$ analogously. The stable subsets containing $p$ are exactly $R(p) \sqcup T_1 \sqcup T_2$, where $T_1 \subseteq S_1$ and $T_2 \subseteq S_2$ are stable. Since $|R(p)|$ is even, exactly $E_1E_2 + O_1O_2$ stable subsets containing $p$ are even, and exactly $E_1O_2 + E_2O_1$ are odd. As $E_1 \ge O_1$ and $E_2 \ge O_2$ by inductive hypothesis, we obtain $E_1E_2 + O_1O_2 \ge E_1O_2 + E_2O_1$ as desired. $\blacksquare$ By the inductive hypothesis, the desired inequality also holds for stable subsets not containing $p$, so we are done.

We proceed by strong induction on \(|S|\), with base case \(S=\varnothing\). The problem is trivial if all points of \(S\) have \(x\)-coordinate 1. Let \(P\in S\) be the point with \(x\)-coordinate 2 and maximum possible \(y\)-coordinate, and let \(A\) be the rectangle with bottom-left corner \((1,1)\) and upper-right corner \(P\). Let \(B\) be the subset of \(S\) above \(A\) and \(C\) the subset of \(S\) to the right of \(C\). Hence \(S=A\sqcup B\sqcup C\). [asy][asy] size(4cm); defaultpen(fontsize(10pt)); draw( (1,0)--(1,3),gray); draw( (2,0)--(2,2),gray); draw( (3,0)--(3,1),gray); draw( (0,1)--(3,1),gray); draw( (0,2)--(2,2),gray); for (int i=3; i<7; i+=1) draw( (0,i)--(1,i),gray); fill( (0,3)--(0,7)--(1,7)--(1,3)--cycle,invisible); draw( (0,3)--(0,7)--(1,7)--(1,3),heavygreen+linewidth(1)); filldraw( (0,0)--(0,3)--(2,3)--(2,0)--cycle,invisible,blue+linewidth(1)); fill( (2,2)--(3,2)--(3,1)--(4,1)--(4,0)--(2,0)--cycle,invisible); draw( (2,2)--(3,2)--(3,1)--(4,1)--(4,0)--(2,0),red+linewidth(1)); real t=.1; filldraw( (1+t,2+t)--(2-t,2+t)--(2-t,3-t)--(1+t,3-t)--cycle,invisible,purple+linewidth(1)); label("\(A\)",(0,1.5),W,blue); label("\(B\)",(0,5),W,heavygreen); label("\(C\)",(3,0),S,red); label("\(P\)",(2,3),NE,purple); [/asy][/asy] Among subsets not containing \(P\), at least half contain an even number of elements by the inductive hypothesis. Focus on subsets containing \(P\) (and thus all of \(A\)). Let \(p_B\ge\frac12\) be the probability a subset of \(B\) contains an even number of elements and \(p_C\ge\frac12\) be the probability a subset of \(C\) contains an even number of elements. Then the probability a subset of \(S\) containing \(P\) contains an even number of elements is \[p_Bp_C+(1-p_B)(1-p_C)=\frac{1+(2p_B-1)(2p_C-1)}2\ge\frac12,\]as desired.

Suppose that $n$ is the largest number such that the $n\text{-th}$ row contains at least one point. For $1\le i \le n$, let $a_i$ be the number of points in the $i \text{-th}$ row. Clearly, $a_1 \ge a_2 \ge \cdots \ge a_n$. In this configuration, define $f(a_1,a_2,\dots,a_n)$ to be the number of stable subsets having an even number of elements minus the number of stable subsets having an odd number of elements (note that inside $f$, there might some $0$'s occasionally, and in that case, you just have to ignore them (of course, $0$'s cannot be between positive integers). Here we are defining $n$ just for convenience and we also (again for convenience) let $f(0,0,\dots)=1$). We want to prove that the values of $f$ are always non-negative. First we claim that for all $n \ge 2$, $$f(a_1,a_2,\dots,a_n) = \sum_{i=0}^{a_n} (-1)^{in} f(a_1-i,a_2-i,\dots,a_{n-1}-i) $$Indeed, if we assume that $i$ elements of the top row are chosen in that stable subset, then we will then have to choose some points from a set of points where in the $i \text{-th}$ row ($1 \le i \le n-1$), there are $a_i - i$ points, and we have already chosen $in$ points, so from here, our claim follows. LEMMA: $f(a_1,a_2,\dots,a_n) \ge 0$ for odd $n$ ; $f(a_1,a_2,\dots,a_n) \ge f(a_1-1,a_2-1,\dots,a_n-1) \ge 0$ for even $n$ PROOF: This is an easy induction on $n$ by using the recurrence. Can someone please check my solution and tell whether it is fine or not

willwin4sure wrote: Let $\mathbb{N}^2$ denote the set of ordered pairs of positive integers. A finite subset $S$ of $\mathbb{N}^2$ is stable if whenever $(x,y)$ is in $S$, then so are all points $(x',y')$ of $\mathbb{N}^2$ with both $x'\leq x$ and $y'\leq y$. Prove that if $S$ is a stable set, then among all stable subsets of $S$ (including the empty set and $S$ itself), at least half of them have an even number of elements. Ashwin Sah and Mehtaab Sawhney Same as few of the above solutions but wadev posting for storage. We induct on $n$, the number of elements in $S$. We give this problem a representation in the cartesian plane for our convenience. We see that the base case $n = 1$ is trivial. Let us say that for all positive integers $k \leq n$, if $S$ has exactly $k$ elements, then the given result holds true. Consider a set $S$ with $n+1$ elements. We impose the cartesian plane representation on the elements of $S$. Here is a sample diagram for $S$ for convenience in explanation. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.298405797293469, xmax = 16.332783342012227, ymin = -3.2076286890721635, ymax = 10.151510615775093; /* image dimensions */ /* draw figures */ fill(shift((3,3)) * rotate(90) * scale(0.1411111111111111) * ((1,0)--expi(2*pi/3)--expi(4*pi/3)--cycle)); /* special point */ fill(shift((6,1)) * rotate(90) * scale(0.1411111111111111) * ((1,0)--expi(2*pi/3)--expi(4*pi/3)--cycle)); /* special point */ fill(shift((5,2)) * rotate(90) * scale(0.1411111111111111) * ((1,0)--expi(2*pi/3)--expi(4*pi/3)--cycle)); /* special point */ fill(shift((1,6)) * rotate(90) * scale(0.1411111111111111) * ((1,0)--expi(2*pi/3)--expi(4*pi/3)--cycle)); /* special point */ /* dots and labels */ dot((0,6),linewidth(4pt) + dotstyle); dot((0,5),linewidth(4pt) + dotstyle); dot((0,4),linewidth(4pt) + dotstyle); dot((0,3),linewidth(4pt) + dotstyle); dot((0,2),linewidth(4pt) + dotstyle); dot((0,1),linewidth(4pt) + dotstyle); dot((0,0),linewidth(4pt) + dotstyle); dot((1,4),linewidth(4pt) + dotstyle); dot((1,3),linewidth(4pt) + dotstyle); dot((1,2),linewidth(4pt) + dotstyle); dot((1,1),linewidth(4pt) + dotstyle); dot((1,0),linewidth(4pt) + dotstyle); dot((2,3),linewidth(4pt) + dotstyle); dot((2,2),linewidth(4pt) + dotstyle); dot((2,1),linewidth(4pt) + dotstyle); dot((2,0),linewidth(4pt) + dotstyle); dot((3,2),linewidth(4pt) + dotstyle); dot((3,1),linewidth(4pt) + dotstyle); dot((3,0),linewidth(4pt) + dotstyle); dot((4,2),linewidth(4pt) + dotstyle); dot((4,1),linewidth(4pt) + dotstyle); dot((4,0),linewidth(4pt) + dotstyle); dot((5,1),linewidth(4pt) + dotstyle); dot((5,0),linewidth(4pt) + dotstyle); dot((6,0),linewidth(4pt) + dotstyle); dot((7,0),linewidth(4pt) + dotstyle); dot((8,0),linewidth(4pt) + dotstyle); dot((2,4),linewidth(4pt) + dotstyle); dot((0,7),linewidth(4pt) + dotstyle); dot((1,5),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\mathbb{S}$ be the set of lattice points whose ordinate and abcissa when written in this order forms an element of the set $S$. We call a lattice point $X \equiv (a, b) \in \mathbb{S}$ as 'even-oriented', if either the abcissa or the ordinate of $X$ is even (that is $2 \mid ab$) and the diagonally adjacent lattice point of $X$ in the north-east direction does not belong to $\mathbb{S}$. The triangle shaped lattice points in the above diagram are few of the even-oriented lattice points in the above diagram. We see that if $\mathbb{S}$ does not contain any even-oriented lattice point, then consider $\mathbb{S}$ with minimal number of lattices points in it (it can be confirmed that $\mathbb{S} \geq 5$ in this case) it must follow that the lattice point with largest abcissa if removed forms a set $\mathbb{S}_1$ with no even-oriented lattice point and lesser number of lattice points than $\mathbb{S}$, contradicting minimality of $\mathbb{S}$, which means that the every such set of lattices points $\mathbb{S}$ must contain an even-oriented point. Consider any even-oriented lattice point $V \in \mathbb{S}$, for our above diagram, we consider $V \equiv (4, 4)$ which is marked by a triangle. We observe that if a stable subset of $\mathbb{S}$ contains $V$, consider sets of lattice points $\mathbb{S}_1$ and $\mathbb{S}_2$ of points such that if $V_1 \in \mathbb{S}_1$ and $V_2 \in \mathbb{S}_2$, then $\text{abcissa}(V_1) > \text{abcissa}(V)$ and $\text{ordinate}(V_2) > \text{ordinate}(V)$. Due to our definition of even-oriented point, it can be confirmed that $S_1 \cap S_2 = \varphi$. We denote $f(S')$ to be the number of stable subsets of a set $S'$ containing even number of elements and let $g(S')$ be the number of stable subsets of $S'$ containing odd number of elements. We see that number of stable subsets of $\mathbb{S}$ containing $V$ and having even number of elements are $f(\mathbb{S}_1)f(\mathbb{S}_2) + g(\mathbb{S}_1)g(\mathbb{S}_2)$ and number of stable subsets of $\mathbb{S}$ containing $V$ and having odd number of elements are $f(\mathbb{S}_1)g(\mathbb{S}_2) + g(\mathbb{S}_1)f(\mathbb{S}_2)$, due to Re-arrangement Inequality, we see that since $f(\mathbb{S}_1) \geq g(\mathbb{S}_1)$ and $f(\mathbb{S}_2) \geq g(\mathbb{S}_2)$ due to our induction hypothesis, we see that $f(\mathbb{S}_1)f(\mathbb{S}_2) + g(\mathbb{S}_1)g(\mathbb{S}_2) \geq f(\mathbb{S}_1)g(\mathbb{S}_2)+f(\mathbb{S}_2)g(\mathbb{S}_1)$. If a stable subset of $\mathbb{S}$ does not contain $V$, then induct and finish the problem. Overall, we get that more than half the number of stable subsets of $\mathbb{S}$ have even number of elements, as desired. A healthy note : We are interested in even-oriented vertices, in my case personally I am interested to look into even-oriented vertices because then number of even-element subsets of $\mathbb{S}$ is $f(\mathbb{S}_1)f(\mathbb{S}_2) + g(\mathbb{S}_1)g(\mathbb{S}_2)$ which strikes thought for Re-arrangement Inequality, here is my motivation to consider such points. I agree the solution is kinda same as official solution but the approach for proof for the existence of even-oriented points are different, that is only reason I posted this solution. Though writing the proof for this lemma properly on a sheet of paper might be troublesome but wadev this is APoS. Also I have solved this problem way back when the TSTST problems were released but I forgot my solution so here we go, new day different approach.

We use strong downwards induction. Define the skin of a stable set $S$ to be the set of all points $(x,y)$ in $S$ such that $(x+1,y+1)$ isn't in $S$. it is easy to see that if the skin of $S$ contains at least two points, one of them must have either $x$ or $y$ coordinate even, and the only stable set with a skin containing a single point is $\{(1,1)\}$. For the base case, we can easily verify that the statement holds if $|S|=0,1$. For the inductive step, suppose $|S| \geq 2$, so the skin of $S$ contains at least two points. Pick some point $P=(x,y)$ in the skin of $S$ with one of its coordinates even. We count the number of stable sets $S' \subseteq S$ based on whether or not they contain $P$: If $P \in S'$, let $R$ denote the rectangle with corners $(1,1)$ and $P=(x,y)$, which contains $xy$ points—an even number. Clearly $R \in S'$, and $S \setminus R$ can be partitioned into two non-adjacent stable subsets $S_1,S_2$. If $S_1$ has $e_1\geq o_1$ stable subsets of even and odd size respectively and $e_2\geq o_2$ are defined similarly (where the inequalities come from inductive hypothesis), then by rearrangement (!) we have $e_1o_1+e_2o_2 \geq e_1o_2+e_2o_1$, so there are at least as many subsets $S'$ with an even number of elements than an odd number of elements. If $P \not \in S'$, then any points above or to the right of $P$ are not in $S'$. Delete these points (including $P$) to get a smaller stable set that $S'$ must be the subset of. By inductive hypothesis there at least as many choices for $S'$ such that $|S'|$ is even than such that $|S'|$ is odd. Combining these two cases, it follows that overall at least half of the choices for $S'$ have an even number of elements, as desired. $\blacksquare$

Same as most solutions above. Use strong induction on $\lvert S\rvert$. The base cases $\lvert S\rvert =0,1$ are trivial. We will prove that if $S_e$ and $S_o$ are the number of even and odd-element stable subsets of $S$, then $f(S)=S_e-S_o\geq 0$. Say that a point $(x,y)\in S$ is dominant if no other $(x',y')\in S$ satisfies $x'> x, y'> y$. Consider dominant $(X,Y)$ from a stable set $S$ to form a new stable $S'$ - choose one of $X,Y$ to be even (which is possible since for any dominant $(x,y)$, we have that one of $(x-1,y), (x,y-1),(x+1,y),(x,y+1)$ must be dominant, and we know $\lvert S\rvert \geq2$. Consider the subset $T\subseteq S$ of lattice points $(x', y')$ satisfying $x'\leq X, y'\leq Y$. Then $S\backslash T$ consists of two (potentially empty) disjoint stable subsets $A, B$. Then, by counting the stable subsets of $S'$ and the stable subsets $M$ of $S$ such that $M\not\subseteq S'$ (which must contain $(X,Y)$): $$f(S)=S_e-S_o=f(S') + A_eB_e + A_oB_o-A_eB_o-A_oB_e=f(S')+(A_e-A_o)(B_e-B_o)=f(S')+f(A)f(B)\geq 0$$by the inductive assumption, completing the inductive step.

Call a subset even if it has an even number of elements and odd, similarly. The case where $|S| = 1$ thus $S = (1, 1)$ is true so we will ignore it. We will work in the coordinate lattice plan with $(x, y)$ matching with its corresponding lattice point. We will do strong induction on $|S|$. Let the barrier of $S$ be the set of points so that $(x + 1, y + 1)$ is not in $S$. Then choose a point $P$ so that one of its coordinates $(x, y)$ are even. Note this clearly exists as there are at least two points on the barrier of $S$ which are adjacent. Then let $R$ be the rectangle with $P$ and $(1, 1)$ as opposite corners, and let $A$ be the region above $P$ and $B$ the region to the right of $P$ so that $A \sqcup B \sqcup R = S$. Let $e_A$ and $e_B$ be the number of even subsets of $A$ and $B$ and $o_A, o_B$ for odd subsets similarly. Then the number of even subsets containing $P$ is $e_Ae_B + o_Ao_B$ and similarly the odd subsets are $e_Ao_B + e_Bo_A$ and by rearrangement inequality we have the number of even subsets being larger than the odd subsets due to $e_A \geq o_A, e_B \geq o_B$(induct. hypothesis). Thus all subsets containing $P$ fit the criteria. If the subset does not contain $P$ then we are done by induction. i wanted to add that this problem gave me an actual headache, I don't like it

Consider a point $P(u,v)$ on the "edge" of our stable set (such that $(u+1,v+1)$ is not in the set) with at least one even coordinate, which can be seen to exist. We use casework on $P$: $P$ is not in the subset: We simply have a smaller stable set. $P$ is in the subset: We must have the $uv$ points in the rectangle with opposite vertices $(0,0)$ and $P$, which is even. The remainder of our set is two smaller stable sets along the $x$- and $y$-axes, which we label $S_1$ and $S_2$. Then we require the inequality \[e_1e_2 + o_1o_2 \ge e_1o_2 + e_2o_1,\]where $e_i$/$o_i$ represent the number of even/odd element subsets of $S_i$. It suffices to have $e_1 \ge o_1$ and $e_2 \ge o_2$ by Rearrangement. Thus we can finish by continuing downwards. $\blacksquare$

If $S$ is the empty set, there is nothing to prove. Otherwise, denoting $X:= \max_{(x,y)\in S}x$ and $Y_i := \max_{(i,y)\in S}y(1 \leq i \leq X)$, it is clear that $S=\{(x, y) | 1\leq x\leq X, 1\leq y\leq Y_x\}$. For every $1\leq x\leq X, 0\leq y\leq X_x$, denote by $a_{x,y}$ the number of stable subsets of $ \{(x',y')|1\leq x' \leq X, (x',y')\in S\}$ containing $(x, y)$(this constraint is omitted for $y=0$) with even cardinality minus the number of such subsets with odd cardinality. For convenience, define $a_{0,y}=1(0\leq y\leq Y)$ and $a_{x,y}=0$ for $(x,y)$ which $a_{x,y}$ is not defined yet. When computing $a_{x,y}$, if $(x,y+1)$ is in the desired subset, then the corresponding value is $a_{x,y+1}$. Otherwise, the subset already contains $y$ elements with x-grid $x$, and the rest is a stable subset consisting of x-grid less than $x$ that contains $(x-1,y)$ . In symbols, we get the recurrence relation $a_{x,y}=a_{x,y+1}+(-1)^ya_{x-1,y}(1\leq x\leq X, 0\leq y\leq X_x)$ . We claim that for any $0\leq x \leq X$, - if $x$ is even, $a_{x, 0}\geq a_{x, 1}\geq \cdots \geq a_{x, Y_x} \geq 0$. - if $x$ is odd, then $a_{x,y}\geq0$ for $y$ even and $a_{x,y}\leq0$ for $y$ odd. The proof goes by induction. The base case $x=0$ is trivial. Supposing the claim holds for $x<X$, - if $x$ is even, $a_{x+1,y}=\sum_{i=y}^{\infty}(-1)^ya_{x,y}$ is an alternating series whose terms' absolute values are non-increasing and eventually becomes zero, so $a_{x+1,y}$ is non-negative if the first term is non-negative($y$ even) and non-positive if the first term is non-positive ($y$ odd) as desired. - if $x$ is odd, all terms in the series $a_{x+1,y}=\sum_{i=y}^{\infty}(-1)^ya_{x,y}$ are positive, so it is immediate that $a_{x+1,y}$ are all positive and do not decrease as $y$ decreases. In particular, $a_{X,0}\geq0$, finishing the proof.

We will denote a set as odd or even depending on its cardinality. We proceed by strong induction on $|S|$. The base cases $|S| = 0$ and $|S| = 1$ are vacuous. For the inductive step, we denote a point $p = (x,y)$ as a boundary point if $(x,y)\in S$, $(x+1, y+1) \notin S$, and at least one of $x,y$ is even. For $|S| \ge 2$, a boundary point obviously exists in $S$ since there must be two adjacent points that satisfy the first two conditions. We will take an arbitrary boundary point $p$, and denote the stable rectangle with upper-right corner $p$ as $R(p)$. Then, partition $S$ into three sets $R(p), A, B$, where $A$ is the set of points above $p$ and $B$ is the set of points to the right of $p$. If the stable subset of $S$ does not contain $p$, we may ignore $p$ and any points above or to the right of $p$. This yields another stable set with less points in it; hence, the inductive hypothesis applies and at least half of the stable subsets not containing $p$ are even. If the stable subset contains $p$, it must contain all of the points in $R(p)$, which is even due to the last condition of being a boundary point. Then, consider $A$ and $B$ as their own stable sets. We let $e_1$ and $o_1$ denote the amount of even and odd subsets of $A$, respectively, and denote $e_2$ and $o_2$ as the amount of even and odd subsets of $B$, respectively. From the inductive hypothesis on $A$ and $B$, we know that $e_1 \ge o_1$ and $e_2 \ge o_2$. The partition causes $R(p), A, B$ to all be independent of each other, so there are $e_1e_2+ o_1o_2$ even stable subsets of $S$ and $e_1o_2+e_2o_1$ odd stable subsets of $S$. Due to the Rearrangment Inequality, we have $e_1e_2+ o_1o_2 \ge e_1o_2+e_2o_1$, implying that the even stable subsets of $S$ that contain $p$ account for at least half of the total stable subsets containing $p$, as desired. $\blacksquare$

Random: the co-authors of this problem alongside UCLA graduate student James Leng recently made a breakthrough on the famous Szemerédi's theorem on arithmetic progressions. You can watch a video about their work here