This is a really nice one. Plot all $(a_i, b_i)$ on the coordinate plane. Center a square with side length $1/\sqrt{i}$ and sides parallel to the coordinate axes at each $(a_i,b_i).$ For any $m > n$ we have $|a_m-a_n|>1/\sqrt{n} > 1/(2\sqrt{n}) + 1/(2\sqrt{m})$ and/or $|b_m-b_n|>1/\sqrt{n} > 1/(2\sqrt{n}) + 1/(2\sqrt{m}).$ Either way, the squares centered at $(a_n,b_n)$ and $(a_m, b_m)$ don't overlap. So none of the squares overlap. But the sum of their areas is the harmonic series which is unbounded, so they are located arbitrarily far from the origin. The end. $\blacksquare$ @below I guess the first thing you have to do is to convince yourself that the answer is negative. I think the idea of "pairing" the sequences comes naturally from the "or" condition. In a way, the pairs of terms have to be "spaced out from each other" for the condition to hold. You want to show that they can't just "clump up" in one region. I think this lends itself naturally to a geometric approach. Once you've plotted them on the coordinate plane, you're going to see that any two points must be a certain distance away from each other in the x direction, or be a certain distance away from each other in the y direction. The square idea is a more workable version of this condition that follows naturally from the idea of wanting to put a "bubble" around each point, forcing the points to spread out. But yes, this is a very hard problem and there's a reason why its P6. I think the fact that no well known techniques are applicable for this problem (at least to my knowledge) would suggest the "out of the box" approach detailed here. I really applaud the proposer.

Any particular reason/motivation to consider plotting $(a_i, b_i)$ on the coordinate plane on the first place? Like, among all of the approaches, why are you motivated to consider that instead of maybe manipulating the sequence which seems much more natural?

I ended up with the same solution, albeit I motivated my solution with number sense. If the $\frac{1}{\sqrt{n}}$ is replaced by $n^{c}$ then $c>-\frac 12$ fails by a block decomposition argument while any $c<-\frac 12$ should work. Therefore the problem must be related to the divergence of harmonic series, so I need to square that thing in some way. If I get too caught up in one sequence and use smth similar to sqrt decomp then I get at most $\int_1^n \sqrt{x}^{-1} dx / \sqrt{n}$ which is bounded by a constant. The 1d version (1 sequence) is easy and uses the same ideas. Then I thought of grouping $(a_i,b_i)$. If I have some form of inequality like $|c_i-c_j|>t$ or something this becomes a 1d problem for a sequence $c_i$. Ideally, $t=O(\frac 1j)$. At this point the 2d interpretation becomes intuitive.

For the sake of completion, here is a construction for $c>-\frac12$. We define these sets as such: $$A_n=\{n^2,n^2+1,\dots,n^2+2n\}$$$$B_n=\{x^2+n | x^2+n<(x+1)^2\}$$For a set of positive integers, $S$, define $m(S)$ the smallest number in $S$. Note that $m(A_n)=n^2$ and $m(B_n)=[\frac{n+1}{2}]^2+n$ Now, we simply Define:(if $n=1$, we define the sum as $0$) $$k\in A_n \Rightarrow a_k=2\sum_{i=1}^{n-1} m(A_i)^c$$$$k\in B_n \Rightarrow b_k=2\sum_{i=1}^{n-1} m(B_i)^c$$Consider the fact that if two numbers $x,y$ are in different $A$ sets, the inequality is correct for $a_x,a_y$ and if they are in different $B$ sets, then it’s correct for $b_x,b_y$ Note that the series converge because : $$2\sum m(A_i)^c = 2\sum n^{2c}$$$$2\sum m(B_i)^c \leq 2\sum (\frac{n+1}{2})^{2c}$$Which we know that for $d<-1$ the series $\sum n^d$ Converges. Now, to finish just note that there doesn’t exist $x,y,n,m$ such that $x,y\in A_n, x,y \in B_m$

Very nice problem with nice idea.

Here's a solution I came up with that doesn't require any ingenuity.

For each $i$, draw a square of side length $1/\sqrt{i}$ with sides parallel to the coordinate axes centered at $(a_i,b_i)$. Then no squares overlap, since for $m>n$ we have $\max\{|a_m-a_n|,|b_m-b_n|\}>1/\sqrt{n}>1/(2\sqrt{n})+1/(2\sqrt{m})$. On the other hand, the sum of the areas of the squares diverges, hence their centers cannot be contained inside a bounded box, which clearly implies the conclusion. $\blacksquare$ Remark: Hm this did not feel very difficult. I guess the main motivation was the following: this condition that "at least one of the two following inequalities holds" is really hard to do anything with, but the condition does imply that $(a_m-a_n)^2+(b_m-b_n)^2>1/n$ for all $m>n$. This lends to an obvious geometric interpretation: if we draw a circle of radius $1/\sqrt{n}$ around each point $(a_n,b_n)$, then any $(a_m,b_m)$ for $m>n$ can't lie inside this circle. But once we get this idea to think about points in the 2d plane it's obvious that we can replace these circles with squares. And then we probably want to have the squares not overlap at all, instead of this weird "points that come after can't lie inside the square", but this turns out to be very easy to do (the idea to take a smaller shape shows up fairly often).

The answer is no; $\{a_n\}$ and $\{b_n\}$ cannot both be bounded. Suppose that $a_i, b_i \in [-N, N]$ for some $N$ and consider the set of points $(a_i, b_i) \in [-N, N]^2$. Observe that we can obviously relax the condition $m > n$. Thus, for any $i$, there cannot exist any points $(a_j, b_j)$ in the square $$S_i = \left[a_i - \frac 1{\sqrt n}, a_i+\frac 1{\sqrt n}\right] \times \left[b_i - \frac 1{\sqrt n}, b_i + \frac 1{\sqrt n}\right].$$(If there was such a point, both $a_j$ and $b_j$ would be within $\frac 1{\sqrt n}$ of $a_i$ and $b_i$.) On the other hand, the area of square $S_i$ is $[S_i] =\frac 4n$. Note that the $S_i$ are non-overlapping and contained in $[-N-1, N+1]^2$. But $$\sum_{i=1}^\infty [S_i] = 4 \sum_{i=1}^\infty \frac 1n$$diverges, contradiction.

Circles >> Squares Note that this is equivalent to the Manhattan distance between $(a_m, b_m)$ and $(a_n, b_n)$. By scaling, this means that we can consider any standard distance metric on ${\mathbb R}^2$, so consider the Euclidean metric. If the two sequences are bounded, then all $(a_k, b_k)$ are in some box. However, each $(a_n, b_n)$ defines a $\pi \cdot \left(\frac{1}{2\sqrt{n}}\right)^2 = \frac{\pi}{4n}$ area circle in which no other circle can intersect. However, this diverges harmonically area-wise, contradiction.

HamstPan38825 wrote: The answer is no; $\{a_n\}$ and $\{b_n\}$ cannot both be bounded. Suppose that $a_i, b_i \in [-N, N]$ for some $N$ and consider the set of points $(a_i, b_i) \in [-N, N]^2$. Observe that we can obviously relax the condition $m > n$. Thus, for any $i$, there cannot exist any points $(a_j, b_j)$ in the square $$S_i = \left[a_i - \frac 1{\sqrt n}, a_i+\frac 1{\sqrt n}\right] \times \left[b_i - \frac 1{\sqrt n}, b_i + \frac 1{\sqrt n}\right].$$(If there was such a point, both $a_j$ and $b_j$ would be within $\frac 1{\sqrt n}$ of $a_i$ and $b_i$.) On the other hand, the area of square $S_i$ is $[S_i] =\frac 4n$. Note that the $S_i$ are non-overlapping and contained in $[-N-1, N+1]^2$. But $$\sum_{i=1}^\infty [S_i] = 4 \sum_{i=1}^\infty \frac 1n$$diverges, contradiction. Idt's Ur squares do intersetc u have taken a larger side length compared to other sols

Very nice problem! 2023-USEMO-2 is related(this is earlier though) The answer is NO. FTSOC, we consider the set of points $S=\{(a_n,b_n)\}_{n=1}^{\infty},$ since $\{a_n\},\{b_n\}$ are both bounded the area of $S$ is finite. However for $m>n,$ the distance of $(a_m,b_m)$ and $(a_n,b_n)$ is at least $$\min\{|a_m-a_n|,|b_m-b_n|\}>\frac{1}{\sqrt n}>\frac{1}{2\sqrt n}+\frac{1}{2\sqrt m}\quad\cdots (*)$$So same as 2023-USEMO-2, take disc $D_1,\ldots ,$ where $D_k=\{(x,y):(x-a_k)^2+(y-b_k)^2<1/{4k}\}.$ Then by $(*)$ we know each 2 disc have no intersection, which gives contradiction because the sum of area $$\sum_{n=1}^{\infty}\pi\cdot\frac 1{4k}\to\infty.\Box$$

We claim that no pair of sequences exist. For the sake of a contradiction, assume otherwise. Then the points $P_i=(a_i, b_i)$ lie on a bounded region $S.$ Without loss of generality, expand this bounded region such that it is a rectangle with sides parallel to the coordinate axes. By our condition, for each point $P_i,$ no other points $P_j$ with $j > i$ are allowed to lie within the square with side-length $\frac{2}{\sqrt{i}}$ centered at $P_i.$ (with sides parallel to the coordinate axes) However, it is clear that we can eliminate the inequality condition. Now, for each square, dilate it by a ratio of $\frac12$ with respect to its center, then it is easy to show that no two squares overlap. Therefore, each point $P_i$ eliminates a region of area at least $\frac{1}{4i}$ within $S.$ (at least one of the quadrants of each square must lie completely within $S$) However, the sum of these regions is the harmonic series but divided by $4$ which is unbounded, therefore $S$ cannot be bounded, a contradiction. QED