Solved with megarnie and OronSH. Let $f(x,y,z):=\sqrt{x}+\sqrt{y}+\sqrt{z}$. Since we want the $f(a_k,b_k,c_k)$'s to be close together, we try partitioning $1,\dots,N$ into "triples" where $a_i=b_j=c_k$, $b_i=c_j=a_k$, and $c_i=a_j=b_k$, so that $f(a_i,b_i,c_i)=f(a_j,b_j,c_j)=f(a_k,b_k,c_k)$. Leftovers become "singles" where $a_k=b_k=c_k$; such singles should satisfy $f(a_k,b_k,c_k)=3\sqrt{a_k}\sim2\sqrt{N}$ or $a_k\sim\tfrac{4}{9}N$. Intuitively, the triples are of the form $(1,N-\epsilon,N-\epsilon)$, $(2,N-\epsilon^+,N-\epsilon^+)$, and so on, where the one small term increases and the two large terms decrease. Eventually, the triples converge toward the singles around the "mean" $\tfrac{4}{9}N$. Because the mean has a "$3$-like" denominator, it reminiscent less of a midpoint and more of a centroid. Armed with that observation, we take a leap of faith: arrange $1,\dots,N$ in an equilateral triangle tiled by equilateral triangles. [asy][asy] size(3cm); draw((-0.5,1.5sqrt(3))--(0.5,1.5sqrt(3))); draw((-1,sqrt(3))--(1,sqrt(3))); draw((-1.5,0.5sqrt(3))--(1.5,0.5sqrt(3))); draw((0,2sqrt(3))--(-2,0)); draw((0.5,1.5sqrt(3))--(-0.5,0.5sqrt(3))); draw((1,sqrt(3))--(0.5,0.5sqrt(3))); draw((0,2sqrt(3))--(2,0)); draw((-0.5,1.5sqrt(3))--(0.5,0.5sqrt(3))); draw((-1,sqrt(3))--(-0.5,0.5sqrt(3))); label("$1$",(0,2.9)); label("$2$",(-0.5,2.03)); label("$3$",(0,2.25)); label("$4$",(0.5,2.03)); label("$5$",(-1,1.16)); label("$6$",(-0.5,1.38)); label("$7$",(0,1.16)); label("$8$",(0.5,1.38)); label("$9$",(1,1.16)); label("$\dots$",(0,0.43)); [/asy][/asy] When $N=n^2$, so that all of $1,\dots,n^2$ fit perfectly, let the triples be the labels of sets of three triangles concentric with the entire triangle, e.g., $(x,y,z):=(1,(n-1)^2+1,n^2)$. Intuitively, it follows that each triple has the same centroid of $\tfrac{4}{9}n^2$. Rigorously, the horizontal cross-sections through the centers of $x$, $y$, and $z$ average out to the cross-section through the centroid, which has length $\tfrac{2}{3}n$; but the former cross-sections are simply $\sqrt{x}+\epsilon$, $\sqrt{y}+\epsilon$, and $\sqrt{z}+\epsilon$, where $|\epsilon|<R=\tfrac{1}{\sqrt{3}}$ (the circumradius of a tile). Therefore, for any concentric triple $(x,y,z)$ we have \begin{align*} \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}+3\epsilon}{3}&=\frac{2}{3}n\\ \implies\left|\sqrt{x}+\sqrt{y}+\sqrt{z}-2\sqrt{N}\right|&\ll2023. \end{align*}If there is a leftover tile in the center, its label is $\lfloor\tfrac{2}{3}n\rfloor^2+\lfloor\tfrac{2}{3}n\rfloor+1$, which can be verified to be a valid single. Finally, when $N=n^2+k$ for $k\le 2n$, remove any $k$ of the $(\tfrac{2\sqrt{N}+2023}{3})^2-(\tfrac{2\sqrt{N}-2023}{3})^2=\tfrac{16184}{9}\sqrt{N}$ valid singles, preferably centered around $\tfrac{4}{9}N$. Accordingly, we have a situation identical to $N=n^2$, except all tiles past the centroid have been shifted up by $k$, resulting in slightly larger $\epsilon$ (but still well below $2023$). $\square$

The idea of the proof is to make a reasonable approximation to the square root. We have the following neat claim. Claim. There exist three permutations $(u_i)_{i=1}^N$, $(v_i)_{i=1}^N$, $(w_i)_{i=1}^{N}$ of the $N = \tfrac{n(n+1)}2$ numbers $1$, $2$, $2$, $3$, $3$, $3$, $\dots$, $n$, $n$, $\dots$, $n$ (there are $i$ occurrences of $i$) such that for all $i$, we have $u_i+v_i+w_i = 2n+1$. Proof. The construction can be phrased inductively, but there is a nice geometrical interpretation: consider an equilateral triangle grid of $\tfrac{n(n+1)}{2}$ points, where numbers in the $i$-th row is labeled $i$. We rotate the equilateral triangles by $120^\circ$, giving three equilateral triangle grids. Then, the three numbers corresponding to the same location in these three triangle grids will form one triple $(a_i, b_i, c_i)$. An example is shown for $N=6$ below. [asy][asy] size(11cm,0); defaultpen(fontsize(9pt)); picture[] pics = {}; picture pic_constr; for(int i=0; i<3; ++i){ picture pic; pics.push(pic); } int n=6; for(int i=0; i<n; ++i){ for(int j=0; j<i+1; ++j){ pair P = ((n-i)/2+j, (n-i)*(sqrt(3)/2)); for(int k=0; k<3; ++k){ dot(pics[k], P); } dot(pic_constr, scale(1.3,1)*P); label(pics[0], (string)(i+1), P, N); label(pics[1], (string)(n-j), P, N); label(pics[2], (string)(n+j-i), P, N); label(pic_constr, "$(" + ((string)(i+1)) + "," + ((string)(n-j)) + "," + ((string)(n+j-i)) + ")$", scale(1.3,1)*P, N); } } for(int i=0; i<3; ++i){ add(pics[i].fit(3cm,3cm), (7*i,0)); } add(pic_constr.fit(6.5cm,5cm), (4.2,-8.5)); [/asy][/asy] This can be easily seen to work by symmetry. $\blacksquare$ To show the problem, suppose $N = \tfrac{n(n+1)}2+m$ for some integers $m\leq n$. The proof proceed in three steps: Take $\boldsymbol m$ numbers away. Let $s(1)$, $\dots$, $s\left(\tfrac{n(n+1)}2\right)$ be the sequence consisting of $1$, $2$, $\dots$, $N$ but with the $m$ numbers $\left\lfloor\tfrac{4N}9\right\rfloor+1$, $\left\lfloor\tfrac{4N}9\right\rfloor+2$, $\dots$, $\left\lfloor\tfrac{4N}9\right\rfloor + m$ removed. For each $i=1,2,\dots,m$, we will make triples $\left(\left\lfloor\tfrac{4N}9\right\rfloor + i, \left\lfloor\tfrac{4N}9\right\rfloor + i, \left\lfloor\tfrac{4N}9\right\rfloor+ i\right)$. This is fine because $$\left| 3\sqrt{\tfrac{4N}9 + m} - 2\sqrt N\right| \leq \left| \sqrt{4N+9n} - 2\sqrt N\right| < 2023.$$Thus, it suffices to triple up $s(1), \dots, s\left(\tfrac{n(n+1)}2\right)$. Next, we claim that for all $i$, we have $|\sqrt{s(i)}-\sqrt i| < 2$. Note that $s(i)=i$ for all $i<4N/9$. For $i>4N/9$, we have $$|\sqrt{s(i)} - \sqrt i| \leq |\sqrt{i+n} - \sqrt i| = \frac{n}{\sqrt{i+n}+\sqrt i} \leq \frac{n}{\tfrac 43\sqrt N} < 2,$$as desired. Approximate square roots with integer. For each $1\leq i\leq N$, we let $t(i)$ be the $i$-th number in the sequence $1 $, $2$, $2$, $2$, $3$, $3$, $3$, $4$, $4$, $4$, $4$,$\dots$. In an explicit formula, we have $t(i) = \left\lfloor \sqrt{2i} + \tfrac 12\right\rfloor$, so we have $$\lvert t(i) - \sqrt{2i}\rvert < \tfrac 12.$$ Apply the claim. By the claim, there exists permutation $(a_i)_{i=1}^N$, $(b_i)_{i=1}^N$ and $(c_i)_{i=1}^N$ satisfying $$t(a_i) + t(b_i) + t(c_i) = 2n+1\quad\text{for all }i.$$For this choice of $a_i$, $b_i$, $c_i$, we use triangle inequality to get that $$\big|\sqrt{a_i} + \sqrt{b_i} + \sqrt{c_i} - 2\sqrt N\big| \leq \frac 1{\sqrt 2}\left( \big| t(a_i) + t(b_i) + t(c_i) - \sqrt{2N}\big| + \frac 32\right) < 4, $$and hence, by using triangle inequality once more, we get that $$|\sqrt{s(a_i)} + \sqrt{s(b_i)} + \sqrt{s(c_i)}-\sqrt{2N}| < 4 + 3\cdot 2 < 2023,$$so there is a way to triple up $s(1)$, $\dots$, $s\left(\tfrac{n(n+1)}2\right)$. From Step 1., the removed $m$ numbers are close to $\tfrac{4N}9$ and forms their own triples, so we are done.

HINT

MarkBcc168 wrote: [asy][asy] size(11cm,0); defaultpen(fontsize(9pt)); picture[] pics = {}; picture pic_constr; for(int i=0; i<3; ++i){ picture pic; pics.push(pic); } int n=6; for(int i=0; i<n; ++i){ for(int j=0; j<i+1; ++j){ pair P = ((n-i)/2+j, (n-i)*(sqrt(3)/2)); for(int k=0; k<3; ++k){ dot(pics[k], P); } dot(pic_constr, scale(1.3,1)*P); label(pics[0], (string)(i+1), P, N); label(pics[1], (string)(n-j), P, N); label(pics[2], (string)(n+j-i), P, N); label(pic_constr, "$(" + ((string)(i+1)) + "," + ((string)(n-j)) + "," + ((string)(n+j-i)) + ")$", scale(1.3,1)*P, N); } } for(int i=0; i<3; ++i){ add(pics[i].fit(3cm,3cm), (7*i,0)); } add(pic_constr.fit(6.5cm,5cm), (4.2,-8.5)); [/asy][/asy] Hence an alternative, more algebraic motivation for the solution is the proof that $\textstyle3\sum k^2=(2n+1)\tfrac{n(n+1)}{2}$, which I find absolutely insane. A month later, the problem still amazes. I wonder if the problem can be generalized to powers $p$ other than $\tfrac{1}{2}$, in the following sense: the mapping from 2D area to 1D cross-section is given by raising to $\tfrac{1}{2}$; perhaps arranging in general simplices may generate different powers? EDIT: Unfortunately, a tetrahedron cannot tile space owing to its dihedral angle of $\arccos\tfrac{1}{3}$, which complicates the most straightforward extension. However, I have my hopes that one can affine transform to a more $\mathbb{Q}\pi$-like structure, in a sense, perhaps considering planar cross-sections of the form $x+y+z=k$ and similar.

How can a problem in SL 2023 be released in Jan 2024?

internationalnick123456 wrote: How can a problem in SL 2023 be released in Jan 2024? What do you mean? It was posted in July...

The actual answer is that countries use the ISL from year $N$ to train their contestants for IMO year $N+1$.