Say a positive integer $n$ is good if there exists a collection of $n$ squares that is tri-connected. If $a$ and $b$ are good, then so is $a+b$ because we may take the union of disjoint copies of two tri-connected sets. Lemma. All good integers are even. Proof. Suppose there is a tri-connected set of $n$ squares. Consider a graph where vertices represent each square and edges are pairs of touching squares. Each vertex has degree $3$, and the sum of the degrees is $3n$, which is also twice the number of edges. Hence $n$ is even. Lemma. $54$ is good. Proof. Consider the following unit consisting of $6$ squares: [asy][asy] size(5cm); pair A=(0, 0); pair B=(1, 0); void buildUnit(... pair[] pts) { for(int i=0; i < pts.length-1; ++i) { pair A=pts[i]; pair B=pts[i+1]; pair C=0.7887*A+0.2113*B; pair D=rotate(-90, C)*A; pair E=rotate(-90, D)*C; pair X1=rotate(210, C)*A; pair X2=X1+C-A; pair F=rotate(90, D)*X1; pair Y1=rotate(90, F)*D; pair G=rotate(-90, C)*X1; pair Y2=X1+X2-Y1; pair A1=X1+X2-A; pair C1=X1+X2-C; pair D1=X1+X2-D; pair F1=X1+X2-F; pair G1=X1+X2-G; filldraw(A--C--D--E--cycle, grey, black); filldraw(D--X1--Y1--F--cycle, grey, black); filldraw(C--X1--Y2--G--cycle, grey, black); filldraw(D1--X2--Y2--F1--cycle, grey, black); filldraw(C1--X2--Y1--G1--cycle, grey, black); filldraw(A1--C1--D1--B--cycle, grey, black); } } buildUnit(A, B); dot(A); dot(B); [/asy][/asy] Each square except the two on the ends are connected to three other squares. By placing $9$ of these units on the sides of a regular nonagon by aligning the marked vertices, we obtain a set of $54$ tri-connected squares: [asy][asy] size(8cm); void buildUnit(... pair[] pts) { for(int i=0; i < pts.length-1; ++i) { pair A=pts[i]; pair B=pts[i+1]; pair C=0.7887*A+0.2113*B; pair D=rotate(-90, C)*A; pair E=rotate(-90, D)*C; pair X1=rotate(210, C)*A; pair X2=X1+C-A; pair F=rotate(90, D)*X1; pair Y1=rotate(90, F)*D; pair G=rotate(-90, C)*X1; pair Y2=X1+X2-Y1; pair A1=X1+X2-A; pair C1=X1+X2-C; pair D1=X1+X2-D; pair F1=X1+X2-F; pair G1=X1+X2-G; filldraw(A--C--D--E--cycle, grey, black); filldraw(D--X1--Y1--F--cycle, grey, black); filldraw(C--X1--Y2--G--cycle, grey, black); filldraw(D1--X2--Y2--F1--cycle, grey, black); filldraw(C1--X2--Y1--G1--cycle, grey, black); filldraw(A1--C1--D1--B--cycle, grey, black); } } for(int i=0; i < 9; ++i) { dot(dir(40*i-90)); buildUnit(dir(40*i-50), dir(40*i-90)); } [/asy][/asy] Lemma. $40$ is good. Proof. Start with a core of four squares arranged in a square pattern, each one connected to two others. On each side, we place three units, making $40$ squares in total: [asy][asy] size(8cm); void buildUnit(... pair[] pts) { for(int i=0; i < pts.length-1; ++i) { pair A=pts[i]; pair B=pts[i+1]; pair C=0.7887*A+0.2113*B; pair D=rotate(-90, C)*A; pair E=rotate(-90, D)*C; pair X1=rotate(210, C)*A; pair X2=X1+C-A; pair F=rotate(90, D)*X1; pair Y1=rotate(90, F)*D; pair G=rotate(-90, C)*X1; pair Y2=X1+X2-Y1; pair A1=X1+X2-A; pair C1=X1+X2-C; pair D1=X1+X2-D; pair F1=X1+X2-F; pair G1=X1+X2-G; filldraw(A--C--D--E--cycle, grey, black); filldraw(D--X1--Y1--F--cycle, grey, black); filldraw(C--X1--Y2--G--cycle, grey, black); filldraw(D1--X2--Y2--F1--cycle, grey, black); filldraw(C1--X2--Y1--G1--cycle, grey, black); filldraw(A1--C1--D1--B--cycle, grey, black); } } pair A=(0.7071, 0); pair B=(1.4142, 0.7071); pair C=(0.7071, 1.4142); pair D=(0, 0.7071); pair E=(-0.7071, 1.4142); pair F=(-1.4142, 0.7071); pair P=(2.3660, 5.8459); pair Q=(-2.3660, 5.8459); pair A1=-A; pair B1=-B; pair C1=-C; pair D1=-D; pair E1=-E; pair F1=-F; pair P1=-P; pair Q1=-Q; filldraw(A--B--C--D--cycle, blue, black); filldraw(D--E--F--A1--cycle, blue, black); filldraw(A1--B1--C1--D1--cycle, blue, black); filldraw(D1--E1--F1--A--cycle, blue, black); buildUnit(E, Q, P, C); buildUnit(E1, Q1, P1, C1); [/asy][/asy] With the previous two lemmas and the observation that the sum of good numbers is good, we see that any number of the form $54m+40n$ is good, where $m$ and $n$ are nonnegative integers. Write this as $2(27m+20n)$. As $\gcd(27, 20)=1$, the largest even integer that cannot be written in this form is $2(27\times 20-27-20)=986$. Consequently, an integer in the interval $[2018, 3018]$ is good if and only if it is even, so there are $501$ good integers in that interval.

We can prove that $6a$ and $8b$ are good for all $a,b>3$. This kill the problem.

Do you need to draw the construction of 54 or 40 tiles accurately in the exam sheet to prove they're possible?

Kayak wrote: Do you need to draw the construction of 54 or 40 tiles accurately in the exam sheet to prove they're possible? No. Just make sure that you accurately describe your construction. Unsolved Extension : What is the largest even $n$ which there does not exists a collection of $n$ squares that is tri-connected?

it turns out that (with help of Geogebra ) every even positive integer $n$ greater than or equal to $28$ is good. First, the "unit" (see post above) can be extended into any unit consisting of $4k+2$ squares for some $k\geq 1$. Now, without loss of generality let the length of the squares be 1. The idea is to create a quadrilateral $ABCD$ if $k$ is divisible by $4$, or pentagon $ABCDE$ if $k$ is not, then make the lengths of the sides and the angles "fit" as to make us able to connect the units in that polygonal form. Let $f(k)$ denote the "length" of the unit consisting of $4k+2$ squares; i.e. in the first two picture, $f(1) = OR$ and $f(2) = H_1 K_1$. It turns out any polygon fits if all the lengths of their sides equals $f(k)$ for some positive integer $k$, and all the angles are greater than $\frac{\pi}{3}$. The third picture shows that there exists a suitable quadrilateral of side lengths $f(2)$, $f(1)$, $f(1)$, and $f(1)$. Using some arguments of increasing the length of the sides, for each $k\geq 1$ there exists a suitable quadrilateral $ABCD$ with $AB = f(k+1)$, $CD = f(k)$, and $BC = DA = f(1)$ (the set of angles are the same for all $k$ if we set the quadrilateral as trapezoid, and the trapezoid form is suitable). Obviously, if $AB = f(k)$ instead of $f(k+1)$, making the quadrilateral a rectangle makes it suitable. From this, we get that $24+4k$ is good for all $k$. Now, we create a regular pentagon $ABCDE$ of side length $f(1)$, which is suitable and implies that $30$ is good. Now, we enlarge the side lengths one by one, increasing the number of squares by $4$ but keeping the pentagon suitable. Since the quadrilateral case is safe, this should be safe as well, which gives us that $30+4k$ is good for all $k\geq 1$. Edit: I think I found a construction for $n = 26$, so every even $n\geq 24$ should be good as well.

Attachments:

The answer is $501$, namely all even numbers. Firstly, by handshaking lemma, $n$ is even. Now we show that there exists a collection of $6n$ tri-connected squares for all $n\geq 4$. 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Construct squares on each of the sides $EA,EH,HB,AB$. Afterwards construct square in the remaining section of $\angle E$ and $\angle B$ such that these two squares are reflections across the midpoint of $BE$. We call such a 6-square pattern a 6-square segment, and the points $N,R$ their endpoints. Construct a regular $n$-gon with side length $NR$ and put a 6-square segment on each side of the regular $6n$-gon then we have a tri-connected collection of $6n$ since obviously the squares won't touch each other. We can do the same thing with the following configuration to show that there exists a collection of $4n$ tri-connect squares for all $4n$ squares. 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Therefore, by Frobenius coin problem, there exists a collection of $n$ tri-connected squares. for each even $2018\leq n\leq 3018$

Answer is all even numbers. We call a triple of squares which are pairwise touching a clover. Then touching two clovers at two points makes a joint. Putting five joints together in a regular pentagon makes a solution with $30$. We can also attach a square to one of the touching points of a clover so that the resulting figure has a line of symmetry and call it a hanging clover. Putting eight of these hanging clovers together gives a solution with $32$. By chicken nuggets this yields everything in the range that is even. If odd, double count incidences to finish.

Funny funny question!! All even work, necessity of evenness is trivial by graph theory. To show all even work, we show that $6n$ and $8n$ are achievable for sufficiently large $n$ (I think $n\geq 4$). To do so, look at the image: to get $6n$ loop the top set of squares and to get $8n$ loop the bottom set of squares.This clearly then finishes by chicken mc nuggets

An illustration as to why $n$ needs to be $\geq 4$ in the above posts [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -271.16906387873996, xmax = 620.6156899608477, ymin = -294.82899422480017, ymax = 230.29861343684163; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((0,0)--(43.30127018922193,25)--(18.30127018922194,68.30127018922192)--(-25,43.30127018922194)--cycle, linewidth(2) + zzttqq); draw((18.30127018922194,68.30127018922192)--(68.30127018922195,68.30127018922191)--(68.30127018922197,118.30127018922192)--(18.301270189221956,118.30127018922194)--cycle, linewidth(2) + zzttqq); draw((68.30127018922195,68.30127018922191)--(43.30127018922193,25)--(86.60254037844382,0)--(111.60254037844385,43.301270189221874)--cycle, linewidth(2) + zzttqq); draw((111.60254037844385,43.301270189221874)--(86.60254037844382,0)--(111.60254037844375,-43.301270189221945)--(161.60254037844373,-43.301270189222)--(186.60254037844376,-1.4210854715202004e-13)--(161.60254037844385,43.3012701892218)--cycle, linewidth(2) + blue); draw((204.90381056766583,118.30127018922173)--(161.60254037844385,93.3012701892218)--(136.60254037844393,136.60254037844376)--(179.9038105676659,161.6025403784437)--cycle, linewidth(2) + zzttqq); draw((136.60254037844393,136.60254037844376)--(111.60254037844388,93.30127018922185)--(68.30127018922197,118.30127018922191)--(93.30127018922202,161.60254037844382)--cycle, linewidth(2) + zzttqq); draw((111.60254037844388,93.30127018922185)--(161.60254037844385,93.3012701892218)--(161.60254037844382,43.30127018922183)--(111.60254037844385,43.30127018922187)--cycle, linewidth(2) + zzttqq); draw((204.90381056766586,118.3012701892217)--(204.90381056766586,68.3012701892217)--(254.9038105676658,68.30127018922168)--(254.90381056766586,118.30127018922167)--cycle, linewidth(2) + zzttqq); draw((254.9038105676658,68.30127018922168)--(229.9038105676658,24.999999999999773)--(273.2050807568877,-2.6290081223123707e-13)--(298.20508075688775,43.301270189221654)--cycle, linewidth(2) + zzttqq); draw((229.9038105676658,24.999999999999773)--(204.90381056766586,68.3012701892217)--(161.6025403784439,43.30127018922178)--(186.60254037844385,-1.5631940186722204e-13)--cycle, linewidth(2) + zzttqq); draw((204.90381056766563,-118.30127018922197)--(204.90381056766572,-68.30127018922197)--(254.90381056766566,-68.30127018922205)--(254.90381056766563,-118.30127018922204)--cycle, linewidth(2) + zzttqq); draw((254.90381056766566,-68.30127018922205)--(229.90381056766574,-25.000000000000103)--(273.2050807568877,-1.508662470066558e-13)--(298.20508075688764,-43.301270189222116)--cycle, linewidth(2) + zzttqq); draw((229.90381056766574,-25.000000000000103)--(204.90381056766572,-68.30127018922197)--(161.60254037844382,-43.30127018922197)--(186.60254037844385,0)--cycle, linewidth(2) + zzttqq); draw((204.9038105676656,-118.30127018922198)--(161.60254037844368,-93.30127018922197)--(136.60254037844368,-136.60254037844385)--(179.90381056766557,-161.6025403784439)--cycle, linewidth(2) + zzttqq); draw((136.60254037844368,-136.60254037844385)--(111.60254037844372,-93.30127018922192)--(68.30127018922177,-118.30127018922187)--(93.30127018922174,-161.60254037844385)--cycle, linewidth(2) + zzttqq); draw((111.60254037844372,-93.30127018922192)--(161.60254037844368,-93.30127018922197)--(161.60254037844376,-43.301270189222)--(111.60254037844379,-43.301270189221945)--cycle, linewidth(2) + zzttqq); draw((1.1368683772161603e-13,0)--(43.30127018922202,-25)--(18.301270189221995,-68.30127018922194)--(-25,-43.30127018922194)--cycle, linewidth(2) + zzttqq); draw((18.301270189221995,-68.30127018922194)--(68.30127018922197,-68.30127018922198)--(68.30127018922195,-118.30127018922198)--(18.301270189221924,-118.30127018922195)--cycle, linewidth(2) + zzttqq); draw((68.30127018922197,-68.30127018922198)--(43.30127018922202,-25)--(86.60254037844392,0)--(111.60254037844388,-43.301270189221995)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw((0,0)--(43.30127018922193,25), linewidth(2) + zzttqq); draw((43.30127018922193,25)--(18.30127018922194,68.30127018922192), linewidth(2) + zzttqq); draw((18.30127018922194,68.30127018922192)--(-25,43.30127018922194), linewidth(2) + zzttqq); draw((-25,43.30127018922194)--(0,0), linewidth(2) + zzttqq); draw((18.30127018922194,68.30127018922192)--(68.30127018922195,68.30127018922191), linewidth(2) + zzttqq); draw((68.30127018922195,68.30127018922191)--(68.30127018922197,118.30127018922192), linewidth(2) + zzttqq); draw((68.30127018922197,118.30127018922192)--(18.301270189221956,118.30127018922194), linewidth(2) + zzttqq); draw((18.301270189221956,118.30127018922194)--(18.30127018922194,68.30127018922192), linewidth(2) + zzttqq); draw((68.30127018922195,68.30127018922191)--(43.30127018922193,25), linewidth(2) + zzttqq); draw((43.30127018922193,25)--(86.60254037844382,0), linewidth(2) + zzttqq); draw((86.60254037844382,0)--(111.60254037844385,43.301270189221874), linewidth(2) + zzttqq); draw((111.60254037844385,43.301270189221874)--(68.30127018922195,68.30127018922191), linewidth(2) + zzttqq); draw((111.60254037844385,43.301270189221874)--(86.60254037844382,0), linewidth(2) + blue); draw((86.60254037844382,0)--(111.60254037844375,-43.301270189221945), linewidth(2) + blue); draw((111.60254037844375,-43.301270189221945)--(161.60254037844373,-43.301270189222), linewidth(2) + blue); draw((161.60254037844373,-43.301270189222)--(186.60254037844376,-1.4210854715202004e-13), linewidth(2) + blue); draw((186.60254037844376,-1.4210854715202004e-13)--(161.60254037844385,43.3012701892218), linewidth(2) + blue); draw((161.60254037844385,43.3012701892218)--(111.60254037844385,43.301270189221874), linewidth(2) + blue); draw((68.30127018922197,118.30127018922192)--(111.60254037844385,43.301270189221874), linewidth(2) + linetype("4 4")); draw((111.60254037844385,43.30127018922187)--(161.60254037844382,43.30127018922183), linewidth(2) + blue); draw((161.60254037844385,43.3012701892218)--(204.90381056766583,118.30127018922173), linewidth(2) + linetype("4 4")); draw((0,0)--(86.60254037844382,0), linewidth(2) + linetype("4 4")); draw((161.60254037844385,43.301270189221796)--(186.60254037844385,-1.5276668818842154e-13), linewidth(2) + blue); draw((273.2050807568877,-2.5579538487363607e-13)--(186.60254037844385,-1.5276668818842154e-13), linewidth(2) + linetype("4 4")); draw((186.60254037844385,-1.5631940186722204e-13)--(161.6025403784439,43.30127018922178), linewidth(2) + blue); draw((186.60254037844376,0)--(161.60254037844376,-43.30127018922199), linewidth(2) + blue); draw((161.60254037844376,-43.30127018922199)--(204.9038105676656,-118.301270189222), linewidth(2) + linetype("4 4")); draw((161.60254037844376,-43.30127018922198)--(186.60254037844385,0), linewidth(2) + blue); draw((186.60254037844385,0)--(161.60254037844382,-43.30127018922197), linewidth(2) + blue); draw((161.60254037844385,-43.301270189221945)--(111.60254037844388,-43.3012701892219), linewidth(2) + blue); draw((68.30127018922187,-118.30127018922192)--(111.60254037844385,-43.30127018922198), linewidth(2) + linetype("4 4")); draw((111.60254037844385,-43.30127018922188)--(161.6025403784438,-43.30127018922197), linewidth(2) + blue); draw((161.6025403784438,-43.30127018922196)--(111.60254037844379,-43.301270189221945), linewidth(2) + blue); draw((111.60254037844379,-43.301270189221945)--(161.60254037844376,-43.301270189222), linewidth(2) + blue); draw((111.6025403784439,-43.30127018922204)--(86.60254037844393,-1.0658141036401503e-13), linewidth(2) + blue); draw((86.60254037844393,0)--(111.60254037844388,-43.3012701892219), linewidth(2) + blue); draw((111.6025403784439,-43.30127018922192)--(86.60254037844393,0), linewidth(2) + blue); draw((86.60254037844393,0)--(111.60254037844388,-43.301270189221995), linewidth(2) + blue); /* dots and labels */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

The example is as beautiful as if the problem was created based on it.