When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

We can prove the assertion with $n$ instead of $2002$ by induction on $n$: Assume it’s proven for $n-1$. If $N\le 2^{n-1}$, then pick $n-1$ elements of the $n$-element set, and perform the construction on those, choosing all the white sets from within the chosen $n-1$-element set. If, on the other hand, $N>2^{n-1}$, then interchange the roles of black and white, and do the same as above, since then the number of black sets would have to be $2^n-N<2^{n-1}$.

is this a plausible solution? I came up with it myself and I think there are some errors in it: First let the set $ S$ be the set of the first 2002 natural numbers. Then define "special" subsets to be subsets that contain an increasing series of consecutive numbers up to 2002, each subset containing all the elements of the previous one: {1}, {1, 2}, {1, 2, 3}, ..., {1, 2, 3, ...., 2002} let the set $ K$ contain all such subsets. Next, define an "awesome subset pair" to be a pair of subsets whose union is in the set $ K$. Now consider parity. For the case when N is even and greater than 3(since the case where N= 0, 1, 2, and 3 are trivial), coloring an awesome subset pair plus its union and the null subset white gets you the smallest such even number of sets N, for which N is greater than 3. A subset in the set $ K$ may have many such "awesome subset pairs" , therefore, for relatively small values of N(e.g. N=4), you use a relatively "small"(as in a subset with little elements) in the set $ K$ and take its "awesome subset pairs". for N=4: {1,2} {} {1} {2} In this way you can keep picking subsets and coloring them white from the set $ K$ and their "awesome subset pairs" for increasing values of N. After the point where N is too large even for all the "awesome subset pairs" and all the subsets in the set $ K$ to hold, we simply use the same technique on the other subsets. First, define a subset to be "weird" if its elements are not consecutive in any way. (e.g. {1, 5, 178, 696}) Next define a series of finite sets $ A_{1}, A_{2}, ..., A_{i}$, for some number i, which each contain subsets which are defined to be "weirdly special", which means that each subset is "weird" and each subset after the first one contains all the elements of the previous one. Let these $ A_i$ sets contain the unions/combinations of the remaining subsets(all of which are "weird"). Then the "awesome subset pair" approach can once again be implemented and now the conditions can be filled for any value of N.

2002 USAMO Problem 1. Let $ S$ be a set with $ 2002$ elements, and let $ N$ be an integer with $ 0 \leq N \leq 2^{2002}$. Prove that it is possible to color every subset of $ S$ either black or white so that the following conditions hold: (a) the union of any two white subsets is white; (b) the union of any two black subsets is black; (c) there are exactly $ N$ white subsets. Official Solution. We prove that this can be done for any $ n$-element set, where $ n$ is a positive integer. Let $ S_{n}=\left \{ 1,2,\ldots,n \right \}$ and an integer $ N$ with $ 0\leq N\leq 2^{n}$. We induct on $ n$. The base case $ n=1$ is trivial. Assume that the desired coloring can be done to the subsets of set $ S_{n}=\left \{ 1,2,\ldots,n \right \}$ and the integer $ N_{n}$ with $ 0\leq N\leq 2^{n}$. We show that there is a desired coloring for set $ S_{n+1}=\left \{ 1,2,\ldots,n,n+1 \right \}$ and integer $ N$ with $ 0\leq N_{n+1}\leq 2^{n+1}$. We consider the following cases. Case (i). $ 0\leq N_{n+1}\leq 2^{n}$. Applying the induction hypothesis to $ S_{n}$ and $ N_{n}=N_{n+1}$, we get a coloring of all subsets of $ S_{n}$ satisfying the given conditions (a), (b) and (c). All uncolored subset of $ S_{n+1}$ contains element $ n+1$, we color all of them blue. It is not hard to see that this coloring of all subsets of $ S_{n+1}$ satisfying the given conditions (a), (b) and (c). Case (ii). $ 2^{n}+1\leq N_{n+1}\leq 2^{n+1}$. Applying the induction hypothesis to $ S_{n}$ and $ N_{n}=2^{n+1} - N_{n+1}$, we get a coloring of all subsets of $ S_{n}$ satisfying the given conditions (a), (b) and (c). All uncolored subset of $ S_{n+1}$ contains element $ n+1$, we color all of them blue. Finally, we switch the color of each subset: if it is blue now, we recolor it red; if it is red now, we recolor it blue. It is not hard to see that this coloring of all subsets of $ S_{n+1}$ satisfying the given conditions (a), (b) and (c). Thus, our induction is complete.

The above solution is nice, but the solution randomly uses the colors blue and red. Here is a solution that is different from all the other solutions presented:

Wow quantumbyte that is an absolutely amazing proof. Did you come up with it? How long did it take you? I knew it wasn't worth reading any of the other solutions.

So I fond the following while doing this problem and I was wondering whether anyone could help me rigorize it:

If this is unclear, please let me know and I'll try to fix it

The solution of @quantumbyte is perfect I will give a proof too which is essentially the same as the official solution. But since I have problems with the phrasing, I would like suggestions

Pretty easy but very nice

Here's my solution: WLOG assume that $S=\{1,2, \dots ,2002\}$. We'll prove a more general result as given below:- GENERALISATION: Let $S(m)=\{1,2, \dots ,m\}$ be a set with $m$ elements, and let $N$ be an integer with $0 \leq N \leq 2^m$. Prove that it is possible to color every subset of $S$ either black or white so that the following conditions hold: the union of any two white subsets is white; the union of any two black subsets is black; there are exactly $N$ white subsets. PROOF: We proceed by induction on $m$. The base cases $m=1$ and $m=2$ are trivially true. So assume that the assertion is true for some $m>2$. We'll show that it is true for $m+1$ also. We use the notation $C(m,N)$ for a coloring which satisfies the given conditions for some $m,N$. Note that the null set $\{\phi \}$ does not affect the problem conditions $1$ and $2$, so we can use this set to increase or decrease the number of white or black colored subsets by one. Our construction will use this fact. Now, according to our inductive hypothesis, $C(m,N)$ exists for all $0 \leq N \leq 2^m$. Call the set $A'=\{a_1,a_2, \dots ,a_r,m+1\}$ the inverse of a subset $A=\{a_1,a_2, \dots ,a_r\}$ of $S(m)$. We make two cases as follows- CASE-1 $[$The null set is colored white in $C(m,N)]$ Color the inverse of a subset of $S(m)$ by the same color as the subset. Then the number of white subsets become $2N$, and we get the coloring $C(m+1,2N)$. Also, if we first discolor the null set and then color it black then we get a new coloring, namely $C(m+1,2N-1)$. CASE-1 $[$The null set is colored black in $C(m,N)]$ Color the inverse of a subset of $S(m)$ by the same color as the subset. Then the number of white subsets become $2N$, and we get the coloring $C(m+1,2N)$. Also, if we first discolor the null set and then color it white then we get a new coloring, namely $C(m+1,2N+1)$. As mentioned before, the color of the null set doesn't affect problem conditions $1$ and $2$, and so we can color the null set accordingly, and easily get $C(m+1,2N)$ and $C(m+1,2N-1)$ for all $0 \leq N \leq 2^m$, which shows that $S(m+1)$ also satisfies the given assertion. Hence, done. $\blacksquare$

WLOG $S={1,2,\dots,2002}$. Define a "subset pyramid" of a set as follows: in the first row, it contains all of the one element subsets, in the second row, it contains all of the two element subsets and so forth such that the kth row contains all of the k element subsets. An example for ${1,2,3,4}$ is shown: $$1,2,3,4$$$$12,13,14,23,24,34$$$$123,234,124,134$$$$1234$$Clearly the union of any two sets is either in the same row or further down the subset pyramid. Also, define a subset pyramid as "inferior" if it has less rows than another subset pyramid. Now, I present an algorithm to construct the white and black sets of subsets. Let's start with $S$ as black. The main claim is that if we take an inferior subset pyramid of another subset pyramid, then we can switch the color and keep the conditions of the problem. This is easy to see: say, for example, we take away the subset pyramid of ${1,2,\dots m}$ from the subset pyramid of $S$, where $m<2002$. We can see that the property of the subset pyramid that the union of two sets are at the same row or further down still holds even after removing the inferior pyramid. Thus, we can take out inferior subset pyramids and switch colors from white to black as much as we need while still satisfying the constraints of the problem. In summary, our algorithm is that we remove an inferior pyramid from $S$, make it white, then take another inferior subset pyramid from that one and make it black, and then with that subset pyramid, we can take an inferior pyramid and make it white and so forth. Let the first inferior pyramid have $a_1$ elements, the next one have $a_2$, and so forth all the way to $a_i$ such that $$0<a_i<\cdots<a_2<a_1\leq 2002$$We have $$N=\sum_{k=1}^{i} (-1)^k*2^{a_k}$$From experimentation with binary representation of base ten numbers, it is easy to see that all $N$ are attainable (each switch in color either adds/replaces with ones or zeroes to the binary representation, so we can handpick each $a_k$ to get our value of $N$). Since we have shown all $N$ are attainable while satisfying the constraints, we are done. Remark: I am really sorry for the lack in clarity of the solution, I am still at a beginning level in trying to put down my ideas on paper, so feel free to ask questions about the solution

Here's a simple constructive solution. Write each subset as a binary string where a $1$ in the one's place represents the first number being in the subset and a $0$ in the one's place represents the first number not being in the subset, and similarly for the other digits. Call the function converting a subset to a number in this way $f$. I claim that coloring the subsets $S$ in which $$f(S)\oplus N\le N$$white and the others black is a valid coloring, where $\oplus$ is the Binary Exclusive-Or operator. This colors exactly $N$ subsets white because The XOR function is reversible. ($f(S)\oplus N\oplus N = f(S)$) It can be seen quite easily that $$f(S_1 \cup S_2)\oplus N = (f(S_1) || f(S_2)) \oplus N$$where $||$ is the bitwise or operator. Now, notice that if both $f(S_1)\oplus N \le N$ and $f(S_2) \oplus N \le N$ then $$(f(S_1) || f(S_2)) \oplus N \le N$$and if both $f(S_1)\oplus N > N$ and $f(S_2) \oplus N > N$ then $$(f(S_1) || f(S_2)) \oplus N > N\blacksquare$$ Of course, further elaboration must be made on the two points above. EDIT: Apparently, $f(S)\oplus N \le N$ is synonymous with what quantumbyte posted, that the maximum element in the subset is in the set that $N$ represents!

Replace 2002 with $n$. We induct on $n$, $n=0$ being trivial. If $N\le 2^{n-1}$, we claim we can use the construction from $n-1$ to get $N$ white subsets, and color the rest black. We are adding the $n$'th element to the construction for $n-1$, call it $x$. The union of any subsets that were originally black is black. The union of a black subset not containing $x$, $S_1$ and a black subset containing $x$, $S_2$, is \[ S_1\cup S_2 = (S_1\cup (S_2\setminus \{x\}))\cup \{x\}, \]which is black. The union of two black subset containing $x$, $S_1$ and $S_2$, is \[ S_1\cup S_2 = ((S_1\setminus \{x\}) \cup (S_2\setminus\{x\}) )\cup \{x\}\]which is black, so this construction works. If $N>2^{n-1}$, we can use the same construction as before, but switching black and white, and replacing $N$ with $2^n-N$.

Solution: it's true for $N=0$, so for $N>0$, let $m,k\in \mathbb{N}_0$ st. $N=2^m+k, k<2^m$ If $k=0,$ color all subsets of $\{1,2,...,m+1\}$ that contains number $m+1$ white. then color other subsets of $\{1,2,...,m+1\}$ which is not taken and have the smallest size until $k=2^m-1$. This algorithm should satisfy all the constraints.

We will generalize $2002$ with $n,$ and induct on $n.$ The base case $n=1$ is trivial. Assume that we can do the task for a $n,$ we will prove that we can then do the task for $n+1.$ For $n+1,$ we have $2^n$ new subsets, other than the previous $2^n$ subsets. For $N= 1, \ldots, 2^n,$ we can just use the induction hypothesis to color the subsets. When $N=2^n + 1,$ what we do is we use symmetry. So, for $2^n+1,$ we color $2^n - 1$ subsets which we previously colored white for the case when $N=2^n-1,$ black; yielding $2^n +1$ white subsets meeting the condition. Essentially, we color $2^n-k$ subsets black(which is obviously possible, ie just make the whites black) where $1 \leq k \leq 2^n$ for $N=2^n + k.$ Therefore, our induction is complete and the conclusion follows.

We will show a more general claim, with every instance of $2002$ in the problem statement replaced with $n$. To show this more general claim, we use induction. The base case $n = 1$ is trivial. For our inductive step, assume that the claim is true for $n$. We will now show the claim for $n+1$, which is equivalent to adding an element $e$ to the set $S$ (we let the set $S$ plus element $e$ be $S_2$). We would like to show that for all $N_2$ from $0$ to $2^{n+1}$, inclusive, there exists a valid coloring of the subsets of $S_2$. Then by inductive hypothesis, we know that for each $N$ from $0$ to $2^n$, inclusive, there exists a coloring of the subsets of $S$ that satisfies the conditions. Take an arbitrary value of $N$. Then let $T$ be the subsets we color white. Consider the set $U$ of subsets, formed by all the subsets in $T$, as well as all the subsets in $T$ plus the element $e$. Then it's easy to see that the union of any two sets in $U$ lies in $U$. Therefore, we have a construction for $N_2 = 2N$, for any arbitrary $N$ from $0$ to $2^n$, inclusive, if we color all the subsets in $U$ white. Now, for odd numbers, we do the following: if the empty set is in $U$, we remove it from $U$, and if the empty set is not in $U$, we add it to $U$. In the first case, we have a construction for $N_2 = 2N-1$, and in the second case, we have a construction for $N_2 = 2N+1$. Moreover, if $N = 2^n$, the empty set has to be in $U$, so $N_2$ never goes out of bounds. This is a valid construction for odd numbers, so our inductive step is complete. $\square$

We will prove the more general case where $|S_n|=n$ for some $n\geq 0$, so that we may use induction. WLOG let the set $S_n$ be $\{1, 2, \dots, n\}$. The base case $n=0$ is trivial. Suppose this is true for $S_{n-1}$; we show it is true for $S_n$. Switch black and white for a subset with $N$ white subsets to get a set with $2^n-N$ white subsets. So, we only need to show this is true for $0\leq N\leq 2^{n-1}$. To achieve this, we may take some coloring in $S_{n-1}$ with $N$ white subsets, and color the missing $2^{n-1}$ subsets containing $n$ black. It is easy to see that this is valid, so we are done. $\blacksquare$

MithsApprentice wrote: Let $S$ be a set with $n$ elements, and let $N$ be an integer with $0 \leq N \leq 2^{n}$. Prove that it is possible to color every subset of $S$ either black or white so that the following conditions hold: (a) the union of any two white subsets is white; (b) the union of any two black subsets is black; (c) there are exactly $N$ white subsets. Induction on $n$;when there are $n+1$ elements in $S$ now.Use the same set as in the $k$th step and colour all remaining subsets black.All these extra "black" subsets are of the form $$\mathcal{M}\cdot n^x +\{n \} \;\;\;\;\;\; x \in \{0,1\}$$,clearly this works since $$\mathcal{M}\cdot n^x+n \cup \mathcal{M}=\mathcal{M}.\blacksquare$$

Let $\mathcal{S}$ have $m$ elements and $\mathcal{P}$ be the set of its subsets. By inducting on $m$ we can show that coloring is possible for any $n \leq |\mathcal{P}|$. If $m = 1$ we color both subsets black for $n = 0$ , the empty set white for $n = 1$ and both subsets white for $n = 2$. Suppose now that a coloring is possible for $m$ (and any $n$). Consider a set $\mathcal{S}$ with $m + 1$ elements. Let $b$ be any element of $\mathcal{S}$. For $n\leq 2^m$ , use induction to color just $n$ subsets of $S - {b}$ white and color black all subsets of $S$ which include $b$. Then the union of two white subsets is still a subset of $S - {b}$ and hence (by assumption) white. The union of two black subsets of $S - {b}$ is black for the same reason. If one black subset includes b, then so does the union, which must therefore be black. For $n > 2^m$, we have $2^{m+1} - n < 2^m$ , so we can find a coloring for $2^{m+1} - n$ and then swap the colors.

We induct on $2^n$, the base case is easy to see. Let the problem be true for $0 \leq N \leq 2^k$. For $N \geq2^k$ the coloring of the induction hypothesis works and for values of $N$ greater than this we just switch white and black.

We will prove this by induction on the number of elements. With 1 element it is trivial. Assume the claim is true for n elements. Either N<2^n or N>2^n. If N<2^n, consider an element "x" and color all subsets containing x black. Then consider the set of elements that aren't x. This problem is equivalent to splitting those into a white and black category with N white subsets. If N>2^n, use a similar argument but replace black with white. Thus when n=2002 the statement is true.

2002 AMO/1 wrote: Let $S$ be a set with 2002 elements, and let $N$ be an integer with $0 \leq N \leq 2^{2002}$. Prove that it is possible to color every subset of $S$ either black or white so that the following conditions hold: (a) the union of any two white subsets is white; (b) the union of any two black subsets is black; (c) there are exactly $N$ white subsets. We will prove this with induction on $m$, the number of elements (in the problem, $m=2002$). Base case. $m=0$. This is trivial, as we can color the empty set either black or white (as desired). Inductive step: Assume it is true for $m=M$. Then, we can generate the following construction for $m=M+1$: If $0 \leq N \leq 2^M$, choose one element and color all subsets with that element black. Color all other elements in accordance with a construction for $m=M$ (with $N$ white subsets). If $2^M \leq N \leq 2^{M+1}$, choose one element and color all subsets with that element white. Color all other elements in accordance with a construction for $m=M$ (with $N-2^M$ white subsets).

Replace $2002$ with $k\ge 1$. We induct on $k$. The result is trivial at $k=1$: for $N=0$, color no subsets white, for $N=1$, color $\{\}$ white, and for $N=2$, color all subsets white. Suppose $k>1$. Let $s$ be an element of $S$ and $S\setminus \{s\} = S'$. If $N\le 2^{k-1}$, use the induction hypothesis to make exactly $N$ white subsets of $S'$, then color every subset of $S$ including $s$ black. If $N>2^{k-1}$, flip the coloring scheme for $N' = 2^k-N \le 2^{k-1}$. We are done.

Generalize: Rephrased Problem wrote: Let $S$ be a set with $n$ elements, and let $N$ be an integer with $0\le N\le2^n$. Prove that it is possible to color every subset of $S$ either black or white so that the following conditions hold: $(\text a)$ the union of any two white subsets is white; $(\text b)$ the union of any two black subsets is black; $(\text c)$ there are exactly $N$ white subsets. We induct on $n$. The base case $n=1$ is true for $N=0$ with all subsets being colored black, and for $N=1$ it's true for only the empty set being white. Suppose it's true for $n$. By switching white subsets with black subsets it suffices to show (for the induction step) that when $S$ has $n+1$ elements, the statement holds for $0\le N\le2^n$. Let $s\in S$ and let $W$ be the set of white subsets of $S\setminus\{s\}$ and let $B$ be the set of black subsets of $S\setminus\{s\}$ for this value of $N$. Then define $W'=\{w\cup\{s\}\mid w\in W\}$ and $B'=\{b\cup\{s\}\mid b\in B\}$. Color the elements of $W'\cup B'\cup B$ black and the elements of $W$ white. This satisfies $(\text a)$ and $(\text c)$. Now we show that it satisfies $(\text b)$. Let $x,y\in W'\cup B'\cup B$. If either of $x,y$ is in $W'\cup B'$, then $s\in x\cup y$ so $x\cup y\notin W$. If $x,y\in B$ then $x\cup y\in B$ too.

We use induction, and it clearly works for $|S|=1$. Assume that it is true for $|S|=k$ elements with the set of subsets $\mathcal{P}(S)$. We now add an element $m$ to $S$ and order $\mathcal{P}(S)$ such that the first $2^k$ elements don't contain $m$, while the last $2^k$ do. By the induction assumption, we may may color any number of $0\le N\le 2^k$ of the first $2^k$ elements white, while we color all the other elements black, and the union of those $N$ white subsets can never form a black subset containing $m$. Also clearly, the union of any black subset stays black firstly by the induction assumption and secondly because for the last $2^k$ subsets the union of two of them will always contain an $m$ and will thus never be white. By symmetry, we could also color all the white subsets black and vice verse, so we can achieve all $0\le N\le 2^k$ number of white subsets such that the conditions hold true.

We prove a stronger statement: Let $S$ be a set with $q$ elements, $\{1,2,\dots, q\}$, and let $N$ be an integer with $0 \leq N \leq 2^q.$ It is possible to color every subset of $S$ either black or white so that the three conditions hold. We will proceed with induction on $q.$ When $q=1$, it is trivial. Assume that it is true for $q=k-1.$ Now, let $q=k.$ If $N\le 2^{k-1}$ then color all subsets containing $k$ black. Note that none of these sets can break the condition (a), because for two sets to have a union that contains $k$, one of the sets must contain $k$, and that set must be colored black. Thus, by the induction hypothesis, we are done, since no matter what $N$ is, there must be a way to color the subsets not containing $k$ so that there are $N$ white subsets. If $N>2^{k-1}$ note that switch the color of every subset and conditions will still be satisfied. The induction step is thus proved, and along with it the assertion.

We induct on $n$ to prove the statement for general $n$; the base case is obvious. Now, first suppose $N \leq 2^{n-1}$. Pick some $n-1$-element subset of $S$, and color its elements such that there are $N$ white subsets according to the conditions given, using the inductive hypothesis. Next, color all subsets containing $N$ black. It is easy to check that this works. For $N > 2^{n-1}$, we can simply re-color every white set black and every black set white because the colors are symmetric.

Replace $2002$ with any positive integer $n$; we then claim that it is possible to color every subset of a set $S_n$ with $n$ elements, say $\{1, 2, \cdots, n\}$ black and white for any $0 \le N \le 2^n$ to satisfy the problem conditions (this obviously finishes the problem). We will prove this with induction, with the base case of $n = 1$ being obvious. Now, for the inductive step, assume that the claim holds for $n = k$; we will then show it holds for $n = k + 1$ as well. First, if $0 \le N \le 2^k$, then we can take the same construction as we did for $S_k$ from the inductive hypothesis, and color any supersets of $\{ k + 1 \}$ black. If $2^k + 1 \le N \le 2^{k + 1}$, then take the construction from $S_k$ where we had to color $2^{k + 1} - N$ subsets white, and color any supersets of $\{k + 1\}$ black. Then, swap the color of every subset; this gives us $N$ white subsets of $S_{k + 1}$ satisfying the problem conditions, and hence the claim is proven.

Let $A(k)$ denote the assertion of the problem statement where $2002$ is replaced with the positive integer $k$. We will show that $A(k)$ is true for all positive integers $k$ via induction on $k$. For a base case, consider $A(1)$. WLOG let $S = \{0\}$. For $N = 0$, let there be no white sets and the rest black, for $N = 1$, let $\emptyset$ be white and $\{0\}$ black, and for $N = 2$, let $\emptyset$ and $\{0\}$ be white and none black. Thus $A(1)$ is true. Now, assume $A(k)$ is true for some fixed $k\ge 1$. We will show that this implies $A(k+1)$ is true as well. Let $T$ be a fixed $k$ element subset of $S$ and let $u$ be the element contaned in $S$ but not $T$. First suppose $N\le 2^k$. Then, since $A(k)$ is true, we can color $N$ subsets of $T$ white and the rest black while satisfying conditions (a) and (b). Then, color all subsets of $S$ containing $u$ black. This coloring clearly satisfies (a) and (b). Now, suppose $N > 2^k$. Then, $2^k - (2^{k+1} - N) < 2^k$, so since $A(k)$ is true, we can color $2^k - (2^{k + 1} - N)$ subsets of $T$ white and the rest black while satisfying conditions (a) and (b). Also color all subsets of $S$ containing $u$ white as well. Similarly to before, this coloring satisfies (a) and (b), and furthermore we get $2^k - (2^{k+1} - N) + 2^k = N$ white subsets, as needed. Hence, $A(k+1)$ is true, so the induction is complete. It follows that $A(2002)$ is true, so we are done.

If $N = 0$ or $2^n$ for some $0\leq n\leq2002$, then the construction is trivial. Let $N = 2^n + k$ with $0<k<2^{n}$. We can then construct $N$ white sets as follows. Let the largest white set be $\{x_1, x_2, \dots, x_{n+1}\}$. Let all subsets of $\{x_1, x_2, \dots, x_n\}$ excluding the empty set be colored white. Then, there are $2^n$ white sets. There exists a maximal $m$ such that \[k = \sum_{i=1}^{m} {n \choose i} + l\]with $0\leq l < {n \choose m+1}$. Then color all sets where $i\leq m$ elements of $\{x_1, \dots, x_n\}$ are removed white. Also color the first $l$ sets where $m+1$ elements are removed white. For example, if $k=n+1$, $\{x_2, x_3, \dots,x_n, x_{n+1}\}, \{x_1, x_3, \dots, x_n, x_{n+1} \}, \dots, \{x_1, x_2,\dots, x_{n-1}, x_{n+1}\}$ and $\{x_3, x_4, \dots, x_{n+1}\}$ are colored white. This construction ensures exactly $N$ subsets are white. Additionally, any union of two white sets either falls into the first $2^n$ subsets or the latter $k$ subsets, i.e. is white. Since all white sets consist of the union of a subset of $\{x_1, x_2, \dots, x_n\}$ and $\{x_{n+1}\}$, they cannot be written as the union of two black sets. Thus, a construction exists for all $N$ and we are done.

This is garbage wording but here we go. Switch $2002$ with $k=n$ we will prove by induction that if the problem works for $k$ than the problem works for $k=n+1$. Suppose we have constructions for $k=n$. If we switch to $k=n+1$ and $N \leq 2^{n}$ we can choose an element color all subsets that contain that element black, for the other subsets repeat a construction. If $N > 2^{n}$ we can make use of a construction for $N'= N-2^n$ Choose an element color the subsets with the element in accordance to the $N'$ construction. Then color the rest of the subsets white. Since this problem works for $k=1$ we are finished.

We will prove this result for all numbers $n$ using induction. For $n=1$, it is clearly possible so this covers our base case. Now, we proceed with the inductive step on $n=k+1$. We have two cases: If $N \le 2^k$, then we take a coloring of subsets of $\{1,2,\dots,k\}$ with $N$ white sets and color every subset containing $k+1$ black. If $N > 2^k$, then we take a coloring of subsets of $\{1,2,\dots,k\}$ with $N-2^k$ white sets and color every subset containing $k+1$ white. Hence, the inductive step is done and we have proved the desired result. $\square$

Replace $2002$ with $n \in \mathbb{N}$ We use induction on $n$ to prove the required result for all natural numbers. Base case: It is trivial for $n = 1$. Inductive hypothesis: Assume for $n = k$, the required result is possible. Inducitve step: For $n = k + 1$. Note that $S$ has $2^{n+1}$ subsets ($2^n$ more when $n =k$), and each element belongs to exactly $2^n$ subsets. Condsider the following cases: Case (i) $0 \leq N \leq 2^n$ Pick an elemet and color al the $2^n$ subsets it belongs to as black. Now by inductive hypothesis, reamining $2^n$ subsets can be colored so that the desired $N$ can be achieved. Case (ii) $2^n < N \leq 2^{n+1}$ Let $ N = 2^n + C$, where $ 1 \leq C \leq 2^n$. Color all the black subsets from case (i) as white to get $2^n$ white subsets and each value of $C$ can be achieved by inductive hypothesis. And we are done. $\square$

Replace $2002$ with some postive integer $n$. We will use induction on $n$ to prove that the result is true for all natural numbers. Base Case: We can see that the base cases $n =1$ and $n=2$ are obvious. Inductive Hypothesis: Assume that the task is possible for $n =k$. Inductive Step: We will prove that the result is true for $n= k+1$. We divide into two cases: If $0\leq N\leq 2^k$, then we take a colouring of subsets of $\{1,2,\dots,k\}$ with $N $white sets and colour every subset containing $k+1$ black. If $2^k < N \leq 2^{k+1}$, then let $N =2^K +C$ where $ 1 \leq C \leq 2^k $. Colour all subset of $\{1,2,\dots,k+1\}$ which contain the element $k+1$ white to get $2^k$ white subsets. Now, we can colour $C$ subsets of $\{1,2,\dots,k\}$ white by our Inductive hypothesis. Thus, we are done .$\blacksquare$ Remark: Here's an example to construct $N=13$ for the set $\{1,2,3,4\}$. Colour all the $2^3$ subsets containing element $4$ white and merge it with a colouring of $5$ white subsets of $\{1,2,,3\}$ from the inductive hypothesis. Thus, we get the following $13$ white subsets: $\{4\},\{1,4\},\{2,4\},\{3,4\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\},\{\phi \},\{3\},\{1,3\},\{2,3\},\{1,2,3\}.$

Replace $2002$ with $k$; we will solve the generalized problem. Proceed with induction on $k$. The base case, $k=1$, is trivial to see. For our inductive hypothesis, assume that it is possible to color a set with $k-1$ elements in such a way where $0\le N\le 2^{k-1}$. Now, let a particular element of the $k$-element set be special; call that element $a$. If $0\le N\le 2^{k-1}$, color all sets containing $a$ black and use the inductive hypothesis to color $N$ of the remaining $2^{k-1}$ subsets (that don't contain $a$) white. If $2^{k-1} < N \le 2^k$, color all $2^{k-1}$ subsets containing $a$ white and use the inductive hypothesis to color $N-2^{k-1}$ of the remaining $2^{k-1}$ subsets (that don't contain $a$) white. Thus done. $\blacksquare$

Consider the binary representation of $N$, including leading zeroes. Starting from $n=2002$, if the $nth$ bit in the binary representation of $N$ is $1$, we color all subsets containing $n$ white. Otherwise, color all subsets containing $n$ black. Then, we repeat this process with $n=2001, n=2001 \ldots n=1$. This works for all $N$ because each number has a unique binary representation. Furthermore, this colors all subsets because each subset has a largest number, say $k$. If the $kth$ bit in $N$ is $1$, it will be white, and if the $kth$ bit in $N$ is $0$, it will be black. Finally, this satisfies the union condition because it is impossible for two black subsets union to be white (this can be seen by considering the largest elements). Similarly, it is impossible for two white subsets union to be black. However, since every subset is colored, the two black subsets union must be black, and vice versa, so we are done.

Replace $2002$ with $n$. Call our set $\{1, \dots, n\}$. We induct on $n$. When $n = 1$, we are done. Say we are done for $n \in \{1, \dots, k - 1\}$. For $n = k$, if $N \leq 2^{k - 1}$, use our inductive hypothesis for the sets whose elements belong to $\{1, \dots k - 1\}$, and colour all sets including $k$ black. This clearly works. For $N \geq 2^{k - 1}$, simply interchange black and white, and finish using our previous case. Hence, we are done.