Disjoint set sum problem. Let us have a set, denoted by $T$, and assign each element a position starting from zero, for e.g. in the set $T=\{1,2,3,4\}$, the positions are $T[0]=1,T[1]=2,T[2]=3,T[3]=4$. Also let's denote total number of elements in a set by $|T|$.and the sum of elements in a set by $S(T)$.
One special property of a set $\{3,4,5,6\}$ is that for two subsets $A,B,\; A\cap B=\emptyset,\; |A||B|\ne0,\;|A|>|B|$, then $ S(A)>S(B)$. As I have analysed, without any proof, all such sets start with a fixed number and are composed of consecutive digits.Why?
Now we need to check that how many pairs of subset can have equal values. For $|T|=4$, the only possibility is $T[0]+T[3]=T[1]+T[2]$ as in the case above out of the total 25 disjoint subset pairs. 
Note/Update: The below analysis is incorrect as we need to select 4 uniformly distributed elements.
This may be counted by $f(4)={}^4{\mathbb C}_4\times 1$ which means choose 4 elements and only one pairing of these can sum equal. Note that $f(4)={}^4{\mathbb C}_4\times f(4)$. Hence we can choose $2k$ elements for 2 disjoint subsets of size $k$ and multiply corresponding arrangement like:
$$f(n)=\sum_{k=2}^{k\le n/2}{}^{n}{\mathbb C}_{2k}\times f(2k)$$
Now with $|T|=5$, it could be written as $f(5)={}^5{\mathbb C}_4\times f(4)$. With $|T|=6$, $f(6)={}^6{\mathbb C}_4\times f(4)+{}^6{\mathbb C}_6\times f(6)$, it can't be used to find $f(6)$. 
Now what could be a possible way to find out the values of $f(n)$?
Note: $f(7)=70$
 A: 
One special property of a set $\{3,4,5,6\}$ is that for two subsets $A,B,A\cap B=\emptyset,|A||B|\neq0,|A|>|B|$, then $S(A)>S(B)$. As I have analysed, without any proof, all such sets start with a fixed number and are composed of consecutive digits. Why?

I'm not sure what you mean by "start with a fixed number" but such sets need not be composed of consecutive digits. As an example consider $\{1,1.1,1.11,1.111\}$.

Let me slightly tweak your notation. Consider a set $T$, $A$ and $B$ be disjoint non-empty subsets of $T$. W.l.o.g., the sets can be treated as lists sorted ascending and take $\text{last}(A) < \text{last}(B)$. By the special property that $S(A) > S(B)$ if $|A| > |B|$, it is clear that we don't need to check whether $S(A) = S(B)$ unless $|A| = |B|$. Consider $|A| = |B| = k$. 
Define the number of pairs of subsets $(A,B)$ that need to be checked as $g(2k)$. We have $g(4) = 1$.
Let $|T|=n$ and $f$ be defined as in the question.
$$f(2k) = \sum_{k=2}^{k\leq n/2} {}^{n}{\mathbb C}_{2k}\cdot g(2k)$$
Now the question is: given $g(4)=1$, how do we find $g(2k), k\geq 3$?

Consider the following set of indices $A' = \{\text{index of } A[i]\text{ in } T \;|\; 0\leq i< k\}$, $B'$ defined similarly. It is clear that if the indices in $B'$ dominate the indices in $A'$, i.e., $B'[i] > A'[i] \;\forall\; 0\leq i< k$, we don't need to check whether $S(A) = S(B)$. It can be shown that this is an "if and only if" condition (see link near the end).
Now, the task is: given a value of $k$, how many pairs of index sets $A'$ and $B'$ can we construct satisfying:


*

*$A'$ and $B'$ are disjoint subsets of $\{0, 1, \ldots, 2k-1\}$ with $|A'| = |B'|$.

*The indices of $B'$ do not all dominate the indices of $A'$.


While I do not know of a counting based approach (solving by hand), I can suggest a programming based one (this is detailed in the link near the end):


*

*Cleverly use for loops to generate pairs of sets satisfying property 1.

*Check for property 2 and increment a count variable if it is satisfied.


As a small check¸ you can do this for $k=3$ by hand and check that $g(6) = 5$, which gives $f(7) = 7\times 5 + \frac{7\times6\times5}{4\times3\times 2}\times 1 = 70$.
Note: This answer is based on the question "Project Euler 106: Necessary and sufficient conditions" and the answer given there.

Example solutions by hand for $k=2, k=3$. $(A',B')$ pairs are written down (1-based numbering) and those with fully dominated indices are crossed out.
$k=2, g(4)=1$: (12,34), (13,24), (23,14)
$k=3, g(6)=5$: (123, 456), (124, 356), (125, 346), (234, 156), (235, 146), (245, 136), (134, 256), (135, 246), (145, 236), (345, 126)
