Let the weights of the stones be $x_1, x_2, \dots, x_{13}$, and $S$ to be the sum of them. We know that $S-x_i$ is even for all $i$ because the rest of the stones can be split up equally. Therefore $x_1 \equiv x_2 \equiv x_3 \equiv \dots \equiv x_{13} \pmod{2}$. Now, consider the set of stones with weights $x_{14}, x_{15}, x_{16}, \dots, x_{26} = \left \lceil \frac{x_1-1}{2} \right \rceil, \left \lceil \frac{x_{2}-1}{2} \right \rceil, \dots, \left \lceil \frac{x_{13}-1}{2} \right \rceil$. (Or $x_{13+k} = \frac{x_k}{2}$ if $x_k$ is even and $\frac{x_k-1}{2}$ if $x_k$ is odd.) Notice that $x_{13+k} < x_{k}$ for all $k$ as long as $x_k>1$. However you think about it, it is simple to see that $x_{14}, x_{15}, \dots, x_{26}$ is a set of stones that satisfies the initial condition. Therefore $x_{14} \equiv x_{15} \equiv \dots \equiv x_{26} \pmod{2}$ just like above. We can repeat the process that we did above for our second set of stones, and then the third, then the fourth, etc. Since $x_k > x_{k+13} > x_{k+26} > \dots$ we must have that all of the values of $x_i$ are $1$ in some set of stones, because the sequence will decrease until then. So, we know that any sequence that satisfies the conditions will go to $(1,1,1,1,1,1,1,1,1,1,1,1,1)$ at some point. We can our way back up from that to get to get our original sequence (by either doing 2x or 2x+1 to every term), and we know that all of the numbers will stay the same the whole time. Therefore $x_1=x_2=\dots=x_{13}$.

parmenides51 wrote: There are $13$ stones each of which weighs an integer number of grams. It is known that any $12$ of them can be put on two pans of a balance scale, six on each pan, so that they are in equilibrium (i.e., each pan will carry an equal total weight). Prove that all stones weigh the same number of grams. Let's label the stones and their weights as $S_1, S_2, S_3, \cdots, S_{12}, S_{13}$. We have $$S_1+S_2+S_3+S_4+S_5+S_6=S_7+S_8+S_9+S_{10}+S_{11}+S_{12}=S$$Now we can replace any one of $S_1$ ~ $S_{12}$ with $S_{13}$, let's say $S_1$. The scale shall remain balanced. $$S_{13}+S_2+S_3+S_4+S_5+S_6=S_7+S_8+S_9+S_{10}+S_{11}+S_{12}=S$$Now $$S_1 = S_{13} = S - (S_2+S_3+S_4+S_5+S_6)$$Similarly, we know that $$S_1 = S_2 =S_3 = \cdots = S_{12} = S_{13}$$

That is not a correct solution @above. You cannot say that $S_{13}+S_2+S_3+S_4+S_5+S_6=S_7+S_8+S_9+S_{10}+S_{11}+S_{12}$, instead you must say that there is a separation of those 12 stones into 2 piles so that the sum is the same. You can’t assume what that separation is.