Consider set $B=\{1,2,\dots , 3k\}, k>1.$ Claim 1:$X_0$ is the largest for set $B.$ We can say by symmetry, $|X_1|=|X_2|.$ And we can show this by induction on $k.$ For base case take $k=2,$ we get $|X_1|=|X_2|=2$ and $|X_3|=3.$ So $X_3 $ is largest. Now if we add $\{4,5,6\}, $ we will get previous set using only $E=\{1,2,3\}$ , sets using only $F=\{4,1,2,3\}$ , sets using only $G=\{5,1,2,3\}$ ,sets using only $H=\{6,1,2,3\}$, sets using only $I=\{4,5,1,2,3\}$, sets using only $J=\{4,6,1,2,3\}$, sets using only $K=\{5,6,1,2,3\}$, sets using only $L=\{4,5,6,1,2,3\}.$ For now sets $F,G,H,I,J,K,L$ will use at least two elements from there sets with $F$ must having $4$, $G$ having $5$, and so on. Now initially we has $X_1=2,X_2=2, X_3=3$ . Now if we see sets, $\{4\},\{5\},\{6\},\{4,5\},\{4,6\}, \{5,6\}.$ Now $X_1=3,X_2=3,X_3=5.$ Now set $E's$ all non empty subset ahs been counted. Now consider set $F.$ Since $F's $ subsets will must have $4,$ it's same as having $E's$ subsets, but everything shifting to $1\mod 3,$ hence it will have $X_1=3,X_2=2,X_3=2.$ Similarly Since $G's $ subsets will must have $5,$ it's same as having $E's$ subsets, but everything shifting to $2\mod 3,$ hence it will have $X_1=2,X_2=3,X_3=2.$ And so on.. adding we get $X_1=20,X_2=20,X_3=23.$ For set $\{1,2,3,4,5,6\}.$ Now for $\{1,2,3,4,5,6,7,8,9\}, $ we will have $X_3$ and infact by this method, we get $|X_3| - |X_1|=2\cdot 3+1=7.$ For $\{1,2,3,4,5,6,7,8,9,10,11,12\}, $ we will have $X_3$ and we get $|X_3| - |X_1|=2\cdot 7+1.$ So this forming recurrence stuff. Clearly $K$ grows more, the difference between $X_3$ and $X_1$ grows very much too. Now consider $C=\{1,2,\cdots,3k,3k+1\},$ It's just the set $B$ but we added ${3k+1}. $ So there are $3$ types of subsets, Subset's of $B$, {3k+1}, D=Subsets must containing $3k+1.$ Claim2: $X_1$ is the largest for $C.$ Now if for $B,$ we have by our claim, $X_1=X_2=a$ and $X_3=b$ and $b>>a.$ Again, note that $D's$ subets are just $ B's$ subsets, but everything shifted with $1\mod 3.$ So $D's$ contribution to $X_1=b,X_2=a,X_3=a$ and the single set $\{3k+1\} $ contributes $1$ to $X_{1}.$ So for set $C,$ we get $X_1=a+b+1,X_2=a+a,X_3=a+b.$ Hence we get $X_1$ to be the largest. Note that $2008=2007+1= 3l+1,$ so by our Claim2, we get $X_1$ to be the largest. And we are done!