Do you mean greatest value $n$?

I think that the answer would be 7 but lemme know if you need to see my work.

Partial Solution Let $L_1, L_2, \dots , L_{10}$ be the number of purple squares in each row, where $\sum_{}{} L_i=n$. Note that we need to have $\sum_{}^{} \binom{L_i}{2} \leq 2\cdot \binom{10}{2}$ otherwise we would have the intersections of two columns and three rows painted purple forming a gremista set. So, by cauchy $$\dfrac{n^2}{10}\leq \sum_{i=1}^{10} L_i^2\leq 180+n\Rightarrow n^2-10n-1800\leq 0\Rightarrow n\leq 47$$And with the following example we prove $44\leq n \leq 47$.

The answer is $46$. We can interpret the columns as sets $A_1,A_2,\dots,A_{10} \subseteq \{1,2,\dots,10\}$ such that $|A_i \cap A_j| \leq 2$ for $1 \leq i<j\leq 10$ and $|A_i \cap A_j \cap A_k|\leq 1$ for $1 \leq i<j<k\leq 10$. Denote $a_i$ by the number of $A_j$'s such that $i \in A_j$ and let $S = \sum_{i=1}^{10} a_i = \sum_{i=1}^{10} |A_i|$. If we sum $|A_i \cap A_j|$ for all $i<j$, then $90 \geq \sum_{i<j} |A_i \cap A_j| = \sum_{i=1}^{10} \binom{a_i}{2} \geq 10\cdot \binom{S/10}{2} \implies S \leq 47$. We are going to show that $S \leq 46$. Case 1: There are at least $43$ pairs $(i,j)$ with $i<j$ such that $|A_i \cap A_j| = 2$. In this case, assume there is some $i$ with $|A_i| \leq 4$. Then for at least seven $j \neq i$, we have $|A_i \cap A_j|=2$, but since $7 > \binom{4}{2}$, this would mean there are distinct $i,j,k$ with $|A_i \cap A_j \cap A_k|=2$, contradiction. Thus, $|A_i| \geq 5$ for all $i$, meaning that $S \geq 50$, contradiction. Case 2: There are at most $41$ pairs $(i,j)$ with $i<j$ such that $|A_i \cap A_j|=2$. In this case, if we sum $|A_i \cap A_j|$ for all $i<j$, we get $86 \geq \sum_{i<j} |A_i \cap A_j| = \sum_{i=1}^{10} \binom{a_i}{2} \geq 10\cdot \binom{S/10}{2} \implies S \leq 46$. Case 3: There are exactly $42$ pairs $(i,j)$ with $i<j$ and $|A_i \cap A_j|=2$. If there exists $i<j$ with $|A_i \cap A_j|=0$, then the argument in case 2 still works. So the remaining three pairs must have one common element pairwise. Thus, we get $87 = \sum_{i=1}^{10} \binom{a_i}{2}$. Assume for contradiction that $S=\sum_{i=1}^{10} a_i = 47$. Let's try to minimize $\sum_{i=1}^{10} \binom{a_i}{2}$. If there are at least $3$ different numbers among the $a_i$'s, then two of the $a_j$'s have the same parity, so making them their average makes this sum smaller. If there are $2$ different numbers among the $a_i$'s, they can't differ by at least $2$ because otherwise we can increment one and decrease the other, getting them closer, making the sum smaller. So there must be $2$ different numbers among the $a_i$'s and they must differ by $1$, let them be $k$ and $k+1$. Then, for some $0 \leq t \leq 10$, $kt + (k+1)(10-t)=47 \implies 47+t=10(k+1) \implies t=3$ and $k=4$, thus $\sum_{i=1}^{10} \binom{a_i}{2}$ is minimized when seven of them are $5$, and three of them are $4$, but in this case $\sum_{i=1}^{10} \binom{a_i}{2} = 7 \cdot \binom{5}{2} + 3\cdot \binom{4}{2} = 88 > 87$, thus we get a contradiction, so $S \leq 46$. An example for $S=46$ is below:

Seems weird a national olympiad would end up asking: "What's the maximum number of edges in a subgraph of $K_{n, n}$ which is $K_{2, 3}$ free?" for some value of $n$ (here $n = 10$).

Here's a solution that has a bit less of casework. Define $n$ and $L_i$ as Rafinha did, and prove that $n \le 47$ in a similar fashion. If $n=47$, consider the following refinement of the inequality: count pairs of colored cells in the same row as usual. Since two rows have colored cells in at most two columns in common, if you fix one row it has at most $9$ pairs of common cells with other rows. This means that if you color $L_i \ge 5$ cells, at least $\binom{L_i}2 - 9$ pairs of columns are not repeated, and we can deduct these pairs from $2\binom{10}2$. Let $x = \max L_i$ and $S = 47-x$ be the total of colored of the remaining rows. Since $47 \ge 4\cdot 10$, $x \ge 5$, and we can use C-S for the other nine rows, for which the sum of binomials is at least $\frac12\left(\frac{S^2}9-S\right) = \frac{S^2-9S}{18}$. With that, \[ \binom x2 + \frac{S^2-9S}{18} \le 90 - \binom x2 + 9 \iff 18x(x-1) + (47-x)(38-x) \le 99 \iff x(19x-103) + 47\cdot 38-99 \le 0 \] This is not true if $x\ge 6$, because $x(19x-103) > 0$ and $47\cdot 38-99 > 0$. Then $x=5$, and so the values of $L_i$ are seven copies of $5$ and three copies of $4$. In this case we deduct $7\cdot \left(\binom 52 - 9\right) = 7$ pairs. However, \[ 7\binom 52 + 3\binom 42 = 88 > 90 - 7, \]a contradiction. This also somehow helps the construction for $n=46$, as it can be shown with a similar work (with a bit more casework) that you need to have six copies of $5$ and four copies of $4$, and the analogous estimate is tight ($84$ pairs). Bear in mind that it must also work for columns as well. One example is: \[ \begin{matrix} \bullet& \bullet& \bullet& \bullet& \bullet& & & & & \\ \bullet& \bullet& & & & \bullet& \bullet& \bullet& & \\ \bullet& & \bullet& & & \bullet& & & \bullet& \bullet\\ \bullet& & & \bullet& & & \bullet& & & \bullet\\ \bullet& & & & \bullet& & & \bullet& \bullet& \\ & \bullet& \bullet& & & & & \bullet& & \bullet\\ & \bullet& & \bullet& & \bullet& & & \bullet& \\ & \bullet& & & \bullet& & \bullet& & \bullet& \bullet\\ & & \bullet& \bullet& & & \bullet& \bullet& \bullet& \\ & & & \bullet& \bullet& \bullet& & \bullet& & \bullet \end{matrix} \]

Twitch Solves ISL solution: The answer is $46$. Call a grid valid if no gremista cells exist. Construction The construction is shown below in the thick green border, which encloses a valid construction of a $10 \times 10$. [asy][asy] void meow(int x, int y) { filldraw(shift( (x%11),-(y%11) )*unitsquare, lightcyan, blue); } for (int y=0; y<11; ++y) { meow(y, y); meow(y+1, y); meow(y+2, y); meow(y+4, y); meow(y+7, y); } draw(box( (1,0), (11,-10) ), deepgreen+2); [/asy][/asy] The reason for the ``extra'' row and column is that this construction is actually obtained by taking an $11 \times 11$ grid and using cells $\{y, y+1, y+2, y+4, y+7\} \pmod{11}$ in the $y$'th row; this gives a valid $11 \times 11$ grid, and deleting a single row and column of a valid $11 \times 11$ will always leave a valid $10 \times 10$. Bound We start with: Claim: There must be at $47$ blue cells in an $10 \times 10$ valid grid. Also, if such a valid grid did exist, it would have to have exactly seven columns with exactly five blue cells each. Proof. Let $a_i$ be the number of blue cells in the $i$th column. Then the usual double-counting argument gives \[ \sum_{1}^{10} \binom{a_i}{2} \le 2 \binom{10}{2} = 90. \]Because $\tbinom{\bullet}{2}$ is convex, for a fixed value of $\sum a_i$ the left-hand side is minimized when any two $a_i$ differ by at most one. So we find that the unique choice of $a_i$ (up to permutation) for which $\sum a_i = 47$ and $\sum \tbinom{a_i}{2} < 90$ is $3\tbinom42 + 7\tbinom52 = 88$; no such choices exist with $\sum a_i \ge 48$. $\blacksquare$ But now we can repeat the same convexity argument on just those seven columns; assuming for contradiction an example for $47$ did exist, let $b_i$ denote the number of blue cells in the $i$th row and in those seven columns. Then we have $\sum b_i = 7 \cdot 5 = 35$ yet \[ \sum_{1}^{10} \binom{b_i}{2} \le 2 \binom 72 = 42 \]but \[ \sum_{1}^{10} \binom{b_i}{2} > 10 \binom{3.5}{2} = 43.75 > 42. \]

We claim that the answer is $46$. Construction: Same as the ones outlined on this thread. Bound: Suppose it is possible to have $47$ colored cells with no gremista. Let the number of blue cells in column $i$ be $a_i$. Then, $a_1+a_2+\cdots+a_{10}=47$. WLOG assume $a_1\geq a_2\geq \cdots \geq a_{10}$. Claim: For any $i\ne j$, we have $a_i+a_j\leq 12$. Proof. Let $a_i\cup a_j$ denote the rows such that at least one of columns $i, j$ and that row have a blue cell together. Define $a_i\cap a_j$ similarly. Then, $10\geq a_i\cup a_j=a_i+a_j-a_i\cap a_j\geq a_i+a_j-2$, which gives $a_i+a_j\leq 12$. $\blacksquare$ Similarly, we get, if $a_i\cup a_j=10$ then $a_k\leq 4$ for evey $k\ne i,j$ because if $a_k\geq 5$ then one of $a_i\cap a_k$ or $a_j\cap a_k$ is at least 3, giving a gremista. We know that $\max\{a_1,a_2,\ldots,a_{10}\}\geq5$. We consider 4 cases. Case 1: $\max\{a_1,a_2,\ldots,a_{10}\}\geq8$. By ordering, $a_1=8$. Then, we have $a_i\leq4$ for every $1<i\leq 10$. Then, $47= a_1+a_2+\cdots+a_{10}\leq8+9\times4=44$, impossible. Case 2: $\max\{a_1,a_2,\ldots,a_{10}\}=7$. By ordering, $a_1=7$. Then, $a_i\leq 5$ for every $1< i\leq 10$. Also, $9a_2\geq a_2+\cdots+a_{10}=47-7=40\Longrightarrow a_2\geq 5$. Hence, $8a_3\geq a_3+a_4+\cdots+a_{10}=35\Longrightarrow a_3\geq 5$. Note that $a_2\cap a_1=2$ is forced. Hence, the other three rows which $a_1$ doesn't occupy must be occupied by $a_2$. But they are also occupied by $a_3$. So, we get a gremista. Case 3: $\max\{a_1,a_2,\ldots,a_{10}\}=6$. By ordering, $a_1=6$. If $a_2=6$ then $a_1\cup a_2=10$ and $a_3, a_4,\ldots, a_{10}\leq 4$. As a result, $47=a_1+a_2+\cdots+a_{10}\leq 44$, impossible. Then, $a_2=a_3=a_4=a_5=a_6=5$. So, $9\geq a_1\cup a_2=a_1+a_2-a_1\cap a_2=6+5-a_1\cap a_2\geq6+5-2=9$ and equality is forced. Let $(1,1)$, $(2,1)$ and $(1,2)$, $(2,2)$ be the common blue cells between $a_1$ and $a_2$, and $(1,10),(2,10)$ both are not blue. Then, for $a_3$, both $(3,1)$ and $(3,2)$ cannot be blue or else we get gremista. Even if one of $(3,1)$, $(3,2)$ are blue then we can have at most two blue cells from $(3,3),\ldots,(3,9)$. So, it means that all blue cells in column 3 are from $(3,3),\ldots,(3,10)$. Then, it forces that $(3,10)$ must be blue. Similarly, $(4,10)$, $(5,10)$ and $(6,10)$ are forced to be blue. Suppose $(3,x)$ and $(1,x)$ is not blue. Then, $(2,x)$, $(4,x)$, $(5,x)$ and $(6,x)$ must be blue from our previous arguement. We get a gremista: $(4,10), (5,10), (6,10), (4,x), (5,x), (6,x)$. Case 4: $\max\{a_1,a_2,\ldots,a_{10}\}=5$. In this case, we get $a_1=a_2=\cdots=a_7=5$. If $a_1\cap a_2=0$ then $a_3\leq 4$. So, in fact, $a_i\cap a_j\geq 1$. The finish is now similar to Case 3 and is hence left as an exercise to the reader

The answer is $\boxed{46}.$ This is achieved by having the $10x10$ lower right corner of the square with the Golomb ruler in each line offset in each consecutive line by $1.$ We show that $47$ is impossible. Note that the sum of the number of blue squares of each column choose $2$ is $\leq 2\binom{10}{2}.$ Therefore with $n=47$ we must have $7$ columns with at least $5$ blue squares. Now, look at only these $7$ columns. There are at least $35$ squares. Thus as the number of blue squares in each row choose $2$ summed is $<2\binom{7}{2}=42.$ However as $\binom{3.5}{2}=4.375,$ as $43.75>42$ we're done$.\blacksquare$