Prove we can remove two objects from each set still keeping the two sets of equal weight 
Let 100 objects of different weights from 1 to 100. We split the
  object set in two sets of equal weight. Prove we can remove two
  objects from each set still keeping the two sets of equal weight.

I tried to generalised by replacing $100$ with something like $100 \rightarrow 4n$ then proving by induction, but without success.
 A: Name the subset containing $1$ as $A$ and the other one as $B$. Choose $m$ the minimal element of $B$. By definition, $1\le m-1\in A$. Choose $n$ the minimal element of $A$ that is bigger than $m$.
$n$ exists as there is no $m$ such that $\frac{(m-1)m}2=25\cdot 101=2525$. If $n>m+1$, then $m,n-1\in B$ and $m-1,n\in A$. So, we're done.
Suppose $n=m+1$. Let $k$ be the minimal number in $B$ that is larger than $m+1$. 
$k$ exists as otherwise $|B|=1$, which is obviuosly impossible. If $k>m+2$, then, $m,k\in B$, $n=m+1,k-1\in A$ gives us the wanted couples.
Suppose $k=n+1=m+2$. Let $l$ be the minimal number in $A$ that is larger than $m+2$.
$l$ exists as $2524$ is not a triangular number.$m,l-1\in B$, $m-1,l\in A$ gives us the wanted couples.
Realize that this proof will work for any $n$, such that $T_n/2$ or $T_n/2-1$ is not triangular. Moreover, if that is the case, counter-examples can be given as 
$$T_n=2T_m\qquad \{1,\ldots,m\},\{m+1,\ldots,n\}$$
$$T_n=2T_m+2\qquad \{1,\ldots,m-2,m-1,m+1\},\{m,m+2,m+3\ldots,n\}$$
As in these cases any couple from $A$ will be less then any couple from $B$.
The conditions can be checked by using the fact that $k=T_n$ for some $n$ iff $8k+1$ is a perfect square and Pell equations. We also need to have $n\equiv -1,0\pmod4$ to be able to split the first set into two equally weighted sets, i.e. for $T_n$ to be even.
