Two is enough. For example take $A = \{502, \dots ,670 \} \cup \{1006, \dots , 1339 \}$, and $B= \{671 , \dots ,1005 \} \cup \{ 1340, 2009 \}$. Check that every pair and every triple from either set sum to either greater than or less than 2010 to see that these are Benelux-sets. Also I think Olympiad problems are supposed to go in the 'unsolved problem' section.

$670 + 1340 = 2010$. Do you want 1339 and $\{1340, \ldots, 2009\}$ instead?

JBL wrote: $670 + 1340 = 2010$. Do you want 1339 and $\{1340, \ldots, 2009\}$ instead? Lol, yeah.

Notice that $502+1509>2010$ thus the subset $A=\{1509,\cdots,2009\}$ is irrelevant. Notice that at most three numbers from the set add up to $2010$ because $502+503+504+505>2010$. Dividing the remaining subset in two parts so each item from $(502+n,1508-n)$ is in different part. $B=\{502,503,\cdots,1005\}$ and $C=\{1006,1007,\cdots,1508\}$ Now lets divide each $B$ and $C$ in two parts and combine them to make no three adition to $2010$ $D=\{502,503,\cdots,753\}$, $E=\{754,755,\cdots,1005\}$, with $B=D \cup E$ $G=\{1006,1007,\cdots,1256\}$, $H=\{1257,1258,\cdots,1508\}$, with $C=G \cup H$ Then $D \cup G$ and $E \cup H$ fullfils the condition.

mszew wrote: Now lets divide each B and C in two parts and combine them to make no three adition to 2010 D=\{502,503,\cdots,753\}, E=\{754,755,\cdots,1005\}, with B=D \cup E G=\{1006,1007,\cdots,1256\}, H=\{1257,1258,\cdots,1508\}, with C=G \cup H Then D \cup G and E \cup H fullfils the condition. $669+670+671 = 2010$ and they are all in $D$... New question, what happens if we add in $501$?

The answer is $n = 2$. Because $n = 1$ is impossible (i.e. $2010 = 502 + 1508$ contradicts the definition of a Benelux-set), we know this is the smallest possible value of $n$. Claim: All bad sets contain $2$ or $3$ elements. Proof. Obviously, a bad set with $1$ element is impossible. If a bad set contains at least $4$ elements, then the minimum possible sum of its elements is $$502 + 503 + 504 + 505 = 2014 > 2010$$contradicting the definition of a bad set. The result follows easily. $\square$ Thus, it's easy to see the elements $\{1509, 1510, 1511, \ldots, 2009\}$ aren't a part of any bad sets, as $2010 - 1509 = 501$ is smaller than all elements in our original set. Thus, we can ignore these integers for the rest of this problem. We also notice $2010 - 502 - 503 = 1005$ is the biggest number appearing in any bad set with $3$ elements (call these bad triplets). Now, consider the $2$ subsets $$\{502, 503, 504, \ldots, 670, 1006, 1007, 1008 \ldots, 1339\};$$$$\{671, 672, 673, \ldots, 1004, 1005, 1340, 1341, 1342, \ldots, 1508\}.$$It's easy to see all pairs of numbers that add to $2010$ are in different sets. (Explicitly, consider the pairs $$(502, 1508), (503, 1507), (504, 1506), \ldots, (1004, 1006)$$which we denote as bad pairs.) Now, notice the only elements from the first subset that could form a \emph{bad triplet} are $$\{502, 503, 504, \ldots, 670\}.$$But $668 + 669 + 670 = 2007 < 2010$, so it's impossible to have a bad triplets only consisting of elements from the first subset. Observe the minimum sum of any $3$ elements from the second subset is $671 + 672 + 673 = 2016 > 2010$, so it's also impossible to have a bad triplets only consisting of elements from the second subset. Thus, we've shown there exists no bad sets only consisting of elements from one of the two subsets, which obviously suffices. $\blacksquare$ Remark: The main idea for this problem is bounding terms that form bad triplets and splitting up terms that form bad pairs (aka symmetry). For some reason, I didn't even consider $n = 2$ when I first began solving this problem (probably because it's "too nice/easy").

$n=1$ is clearly impossible as $502+1508=2010$, so we claim $n=2$. For $n=2$, take the sets: $$\{502,\ldots,670,1006,\ldots,1339\}\text{ and }\{671,\ldots,1005,1340,\ldots,2009\}.$$