An example with $1006$ integers is $M = \{1016,1017,\ldots,2021\}$ which works since $a+b-c \geq 1016 + 1016 - 2021 = 11$ for any $a,b,c \in M$. Now suppose there is $M$ with the given property and at least $1007$ elements. Note that there are at least $1006$ negative integers of the form $a-b$ for distinct $a,b\in M$ - indeed, if the elements of $M$ are $x_1 < x_2 < \ldots < x_{1007} < \ldots$ then $x_i - x_{1007}$, $i=1,\ldots,1006$ are such. Colour these differences in blue. Also, colour each element of $-M$ (i.e. the integers $-x_1, -x_2,\ldots, -x_{1007}, \ldots$) in red. Now there are two cases: - If there is an integer which is both red and blue, then $a - b = -x_i$ for some $a,b,x_i \in M$ and so we have $a,b,c \in M$ which $a+b-c = 0$, which contradicts the hypothesis. - Otherwise, no integer is both red and blue and since all coloured integers (which are at least $2013$ in total) are between $(-2020)$ and $(-1)$, it follows that there is a difference of a red and a blue integer equal to $m\leq 10$ and we get a contradiction as in the previous case (but with $0$ replaced by $m$).

Solution (by Contestant ROU 4 @CezarOfArithmetics). An example of a set with $1006$ elements has already been provided above. We now give a not really different (similar in flavour, different phrasing) proof of the bound. Further assume that $M$ has (at least) $1007$ elements and let $A$ denote the given set. Let $a_{1}<a_{2}<\dots<a_{1007}$ be the elements of $M$. Consider the following set (it's easy to check that it's well-defined and also a subset of $A$) $$M'=\left\{a_{1007}-a_{1006}, a_{1007}-a_{1005},\dots, a_{1007}-a_{1}, a_{1007}-a_{1}+1, a_{1007}-a_{1}+2,\dots, a_{1007}-a_{1}+10\right\}.$$ By the problem's condition, $M$ and $M'$ must be disjoint. But since $|M|+|M'|=1007+1016=2023>|A|=2021$ and $M, M'\subset A, M\cap M'=\emptyset$, we get a contradiction. This ends our proof.

trigadd123 wrote: Solution (by Contestant ROU 4 @CezarOfArithmetics). An example of a set with $1006$ elements has already been provided above. We now give a not really different (similar in flavour, different phrasing) proof of the bound. Further assume that $M$ has (at least) $1007$ elements and let $A$ denote the given set. Let $a_{1}<a_{2}<\dots<a_{1007}$ be the elements of $M$. Consider the following set (it's easy to check that it's well-defined and also a subset of $A$) $$M'=\left\{a_{1007}-a_{1006}, a_{1007}-a_{1005},\dots, a_{1007}-a_{1}, a_{1007}-a_{1}+1, a_{1007}-a_{1}+2,\dots, a_{1007}-a_{1}+10\right\}.$$ By the problem's condition, $M$ and $M'$ must be disjoint. But since $|M|+|M'|=1007+1016=2023>|A|=2021$ and $M, M'\subset A, M\cap M'=\emptyset$, we get a contradiction. This ends our proof. Do you know the expected scores of the Romania team and if so can you please share them?

Can someone check if this solution is correct? The answer is $1006$ and the example is already shown above. Now let's show that $M$ can't have more than $1006$ elements. If we pick $a=b=c=k$ for an element $k$ in $M$ we get that $k>10$, so each of the elements in $M$ is at least $11$. Now let the largest element in $M$ be $m$. If $j\leq \lfloor \frac{m-10}{2}\rfloor$ is a positive integer we can't have both of $10+j$ and $m-j$ in $M$, because then if we choose $a=10+j ; b=m-j ;c=m$ we get $|a+b-c|=10>10$, impossible. Now we get that from each of the pairs: $11$ and $m-1$; $12$ and $m-2$ ; ... ;$10+\lfloor \frac{m-10}{2}\rfloor$ and $m-\lfloor \frac{m-10}{2}\rfloor$ we can have at most one of the numbers in $M$ $\Rightarrow$ Except $m$ we can have at most $\lfloor \frac{m-10}{2}\rfloor$ other numbers in $M$, thus the largest possible number of elements is $1+\lfloor \frac{m-10}{2}\rfloor=\lfloor \frac{m-8}{2}\rfloor$, which is $1006$ when $m=2021$ and less(or equal) when $m<2021$, so the desired maximum is indeed $1006$. Remark:$10+\lfloor \frac{m-10}{2}\rfloor$ and $m-\lfloor \frac{m-10}{2}\rfloor$ can be equal when $m$ is even, but that doesn't change the desired bound. When $m$ is odd they differ by $1$, thus the pairs that I listed contain all possible numbers in $M$(we can't have the numbers that are larger than $m$ and the numbers from 1 to 10)

Pitagar wrote: Can someone check if this solution is correct? The answer is $1006$ and the example is already shown above. Now let's show that $M$ can't have more than $1006$ elements. If we pick $a=b=c=k$ for an element $k$ in $M$ we get that $k>10$, so each of the elements in $M$ is at least $11$. Now let the largest element in $M$ be $m$. If $j\leq \lfloor \frac{m-10}{2}\rfloor$ is a positive integer we can't have both of $10+j$ and $m-j$ in $M$, because then if we choose $a=10+j ; b=m-j ;c=m$ we get $|a+b-c|=10>10$, impossible. Now we get that from each of the pairs: $11$ and $m-1$; $12$ and $m-2$ ; ... ;$10+\lfloor \frac{m-10}{2}\rfloor$ and $m-\lfloor \frac{m-10}{2}\rfloor$ we can have at most one of the numbers in $M$ $\Rightarrow$ Except $m$ we can have at most $\lfloor \frac{m-10}{2}\rfloor$ other numbers in $M$, thus the largest possible number of elements is $1+\lfloor \frac{m-10}{2}\rfloor=\lfloor \frac{m-8}{2}\rfloor$, which is $1006$ when $m=2021$ and less(or equal) when $m<2021$, so the desired maximum is indeed $1006$. Remark:$10+\lfloor \frac{m-10}{2}\rfloor$ and $m-\lfloor \frac{m-10}{2}\rfloor$ can be equal when $m$ is even, but that doesn't change the desired bound. When $m$ is odd they differ by $1$, thus the pairs that I listed contain all possible numbers in $M$(we can't have the numbers that are larger than $m$ and the numbers from 1 to 10) Yes, it is true.

trigadd123 wrote: Solution (by Contestant ROU 4 @CezarOfArithmetics). An example of a set with $1006$ elements has already been provided above. We now give a not really different (similar in flavour, different phrasing) proof of the bound. Further assume that $M$ has (at least) $1007$ elements and let $A$ denote the given set. Let $a_{1}<a_{2}<\dots<a_{1007}$ be the elements of $M$. Consider the following set (it's easy to check that it's well-defined and also a subset of $A$) $$M'=\left\{a_{1007}-a_{1006}, a_{1007}-a_{1005},\dots, a_{1007}-a_{1}, a_{1007}-a_{1}+1, a_{1007}-a_{1}+2,\dots, a_{1007}-a_{1}+10\right\}.$$ By the problem's condition, $M$ and $M'$ must be disjoint. But since $|M|+|M'|=1007+1016=2023>|A|=2021$ and $M, M'\subset A, M\cap M'=\emptyset$, we get a contradiction. This ends our proof. sry im sort of confused but how do we know that $a_{1007}-a_{1}+10$ is an element of $\left\{1,2,...,2021\right\}$? isnt $a_1 = 1$ and $a_{1007} = 2021$ a counter example?

@above Check that all elements of $M$ must be higher than or equal to $11$ (otherwise, if for some $k$ we had $a_{k}\leq 10$, we'd get $|a_k-a_k+a_k|\leq10$, which is a contradiction). Then it immediately follows that $M$ really is well-defined. Indeed, I should not have let this slip (but I pretty much just wanted to highlight the core idea).

Could someone check if this works please? A construction for $M$ with $1006$ elements has already been given. Assume there exists an $M$ with $1007$ elements. Let the elements of $M$ be $a_1, a_2, a_3 ... a_{1007}$ with $a_1 < a_2 < a_3 ...< a_{1007}$. Claim 1: $a_1 \leq 1005$ We must have one of the two following cases hold $2a_1 - a_{1007} > 10$ or $2a_1 - a_{1007} < -10$. If the first were to hold we would have $a_1 > 10 + a_{1007} - a_1 \geq 10 + 1006 = 1016$. However, $a_1 > 1016$ is impossible as it forces $a_{1007} > 2022$. So we must have the second inequality holding which gives $2a_1 < a_{1007} - 10 \leq 2021 - 10 = 2011 \Rightarrow a_1 \leq 1005$ Claim 2: $a_1 \geq 508$ Assume the contrary and that $a_1 \leq 507$. Let $a_n$ be the largest element in $M$ such that $a_n \leq 2011 - a_1$. Let $a_k \leq a_n$. We know that $a_k + a_1 + 10 \leq a_n + a_1 + 10 \leq 2021$. Thus, the range of $i$ satisfying $a_k + a_1 - 10 \leq i \leq a_k + a_1 + 10$ lies completely within $1 \leq i \leq 2021$. By the given condition we can't have $a_k + a_1 - 10 \leq a_i \leq a_k + a_1 + 10$ and so the range $[2a_1 - 10, 2a_1 + 10]$ prohibits $21$ integers. For any $a_k$ that we add on the added number of integers that we prohibit from being used, that is the range $[a_1 + a_k - 10, a_1 + a_k + 10]$, is greater than $2$. And so we can obtain a lower bound for the number of prohibited integers, $P$, which is $21 + 2(n - 1) \leq P$. Since $a_n \leq 1014 + n$ we obtain that all $n \leq 997 - a_1$ meets our inequality $a_n \leq 2011 - a_1$. So we can adjust our lower bound to $21 + 2(997 - a_1 - 1) = 2013 - 2a_1 \leq P$. However, we must have $P + 1007 \leq a_n - a_1 + 1$ but this implies $2013 - 2a_1 + 1007 \leq a_n - a_1 + 1 \Rightarrow 3019 \leq a_n + a_1 \leq 2021 + a_1 \Rightarrow 998 \leq a_1$ a contradiction. Claim 3: We can't have $508 \leq a_1 \leq 1005$ Assume the contrary and that $508 \leq a_1 \leq 1005$. Let $a_n$ be the largest term such that $a_n \leq 1005$. Consider the integers that the range $[a_1 + a_k - 10, a_1 + a_k + 10]$ prohibits from being used where $508 \leq a_1 \leq a_k \leq a_n \leq 1005$. $a_1 + a_k - 10 \geq 2a_1 - 10 \geq 1006$ and $a_1 + a_k + 10 \leq 2a_n + 10 \leq 2020$ and thus the integers being prohibited from use all lie within the range $[1006, 2020]$. Similar to what we do in Claim $2$, we have that $P \geq 21 + 2(n - 1)$ where $P$ is the number of integers being prohibited. We know we have $n$ terms less than or equal to $1005$ and we have a maximum of $1016 - P$ terms in the range $[1006, 2021]$. So we must have $n + 1016 - P = 1007$. However, inputting our lower bound of $P$ we get $n + 1015 - P \leq n + 1015 - (21 + 2n - 2) = 996 - n \leq 995$ which is a contradiction. This proves Claim $3$ but this violates our results from Claim $1$ and $2$ forcing a contradiction and so creating such an $M$ is impossible.

Here is a cleaner version of Pitagar's amazing solution! An example with 1006 numbers is $M = \{1016,1017,\ldots,2021\}$, which works since $a+b-c \geq 1016+1016-2021 = 11$ for any $a,b,c$ from $M$. We move on to the bound. If $k$ is the smallest element of $M$, then $k=|k+k-k|\geq 11$. Let $m\leq 2021$ be the largest element of $M$. If $1\leq j \leq \left\lfloor\frac{m-10}{2}\right\rfloor$, then we cannot have $10 + j$ and $m-j$ simultaneously in $M$, otherwise $a = 10 + j, b = m - j, c = m$ would give $a+b-c = 10$. Hence $M$ contains at most $\left\lfloor\frac{m-10}{2}\right\rfloor \leq 1005$ numbers different from $m$, and the result follows.

Putting $a=b=c$ yields that $M\subset [11..2021].$ It boils down to show that for any $M\subset [11..2021], |M|=1007$ there exist $ a,b,c\in M$ with $ |a+b-c|\le 10.$ The following claim with proper adjustment was used repeatedly in BMO 2023, p4. Claim. For any set $ M$ of $ 1007$ (or more) integers in $ [11,m], m\le 2021,$ there exist $ a,b\in M$ with $ a+b=m+10.$ Proof. Let's prove it for $ m=2021$ to avoid some casework; $ m$ odd and $ m$ even have to be considered. I just want to present the idea. Assume on the contrary that it is not true. Consider the pairs $ (1015-i, 1016+i, i=0,1,\dots,1004).$ In each of these pairs there is only one element of $ M.$ (otherwise the sum of the numbers of that pair would be 2031). This means that the numbers in $ M$ that are up to $ 2020$ are at most $ 1005.$ Hence, $ |M|\le 1006,$ contradiction. $ \square$ Assume now, searching for contradiction, that for any $ a,b,c\in M$ we have $ |a+b-c|>10.$ Applying the Claim, we get $ a,b\in M$ with $ a+b=2031.$ It means $ 2021\not \in M.$ So, we have a set $ M$ with $ 1007$ elements all of them in $ [11,2020].$ We apply the same Claim but with $ 2020$ instead of $ 2021.$ Analogously, we get $ a,b\in M$ with $ a+b=2030,$ which means $ 2020\not\in M$ and so on. Thus, in the end, we get that no integer in $ [11,2021]$ belongs to $ M,$ contradiction. Remark. I happened to see this problem after seeing BMO 2023, p4. The same idea used there. Some more comments can be found on my blog.

Solved with mathmax12 We claim that the answer is $\boxed{1006}$, note that $M = {1016,1017,1018, ......... 2021}$, now we do a proof by contradiction, so if it works then there are at least $1006$ negative numbers in the form of $a-b$,(we could not continue from here)