Claim: It is always possible to equate all numbers iff $ n$ is even. First we prove that for all odd $ n$, there exists some $ n$-tuple such that it's impossible to equate all numbers. Consider the $ n$-tuple $ (2,2,\cdots ,2,2,1)$ (total odd numbers) Suppose to the contrary that it is possible to yield equal numbers. i.e. $ m,m,\cdots ,m,m,m$ Clearly $ m>2$ Note that if the operation $ (a,b)\to (a+b,a+b)$ can get $ (2,2,\cdots ,2,2,1)\to(m,m,\cdots ,m,m,m)$ Then the reverse operation $ (p,p)\to (q,p-q)$ can get $ (m,m,\cdots ,m,m,m)\to(2,2,\cdots ,2,2,1)$ Note that the operation $ (p,p)\to (q,p-q)$ decreases both numbers. The operation must act on each $ m$ in the set $ (m,m,\cdots ,m,m,m)$ Also this reverse operation requires both numbers to be equal. Thus it will always act on the $ m$s in pairs. But there are odd number of $ m$s. Hence there will remain one $ m$ on which this reverse operation never acts. Contradiction. Now it remains to prove that for all even $ n$ every $ n$-tuple can be transformed into equal numbers. Lets treat the $ n$-tuple as a $ m$-tuple of pair of elements of the $ n$-tuple. ($ n=2m$) That is $ (a_1,a_2,\cdots ,a_n)=((a_1,a_2),(a_3,a_4),\cdots ,(a_{2m-1},a_{2m}))$ Lets do the operation on the 2 elements of each pair. Let $ b_i=(a_{2i-1}+a_{2i},a_{2i-1}+a_{2i})$ $ ((a_1,a_2),(a_3,a_4),\cdots ,(a_{2m-1},a_{2m}))\to (b_1,b_2,\cdots ,b_m)$ (From now the two elements of each pair will never be different) Suppose $ b_i=(a,a),b_j=(b,b)$ $ (b_i,b_j)\to (b_i+b_j,b_i+b_j)$ is equivalent to $ ((a,a),(b,b))\to ((a+b,a+b),(a+b,a+b))$. which is the composition of 2 original operations. Actually now we treat the $ n$-tuple as an $ m$-tuple. also note that we now have a extra operation: we can multiply some $ b_i$ by $ 2$ whenever we like, because $ (a,a)\to (2a,2a)$ Thus at some point we can multiply powers of $ 2$ to some $ b_i$s so that the largest power of 2 dividing $ b_i$ are same for all $ b_i$ Let $ o(x)$ denote the largest odd divisor of $ x$. If $ b_i\not =b_j$ have the same largest power of $ 2$ then note that $ o(b_i+b_j)<\max(o(b_i),o(b_j))$ (It's not hard to see, because the sum of odd numbers is even) At some particular state of the blackboard define $ l,s$ such that for all $ i$, $ o(b_l)\ge o(b_i)\ge o(b_s)$ Now the strategy of obtaining all equal numbers is: $ \boxed{}$ operate pairs of the $ n$-tuple and make equal to get it into $ (b_1,b_2,\cdots b_m)$ form. $ \boxed{}$ get all numbers into equal largest powers of $ 2$ $ \boxed{}$ operate on $ (b_l,b_s)$ $ \boxed{}$ get all numbers into equal largest powers of $ 2$ $ \boxed{}$ operate on $ (b_l,b_s)$ $ \boxed{}$ get all numbers into equal largest powers of $ 2$ $ \boxed{}$ operate on $ (b_l,b_s)$ $ \cdots$ Now it remains to prove that this strategy will yield equal numbers within finite number of steps. After all "operate on $ (b_l,b_s)$" steps, the-largest-odd-number-dividing-at-least-one-$ b_i$ decrease unless $ o(b_l)=o(b_s)$ But the-largest-odd-number-dividing-at-least-one-$ b_i$ can't decrease indefinitely. Hence, after finite number of steps, $ o(b_l)=o(b_s)$ But $ o(b_l)=o(b_s)$ implies $ o(b_i)$ is same for all $ i$ And then we can equate the largest power of $ 2$ of all $ b_i$ Now we have all equal numbers. Example: $ 3,10,7,9,4,6$ $ 13,13,16,16,10,10$ $ 26,26,16,16,20,20$ $ \cdots$ $ 2^4\cdot 13,2^4\cdot 13,2^4\cdot 1,2^4\cdot 1,2^4\cdot 5,2^4\cdot 5$ $ 2^4\cdot 14,2^4\cdot 14,2^4\cdot 14,2^4\cdot 14,2^4\cdot 5,2^4\cdot 5$ (from now there won't be any unnecessary repeatations.) $ 2^5\cdot 7,2^5\cdot 7,2^5\cdot 5$ $ 2^5\cdot 7,2^5\cdot 12,2^5\cdot 12$ $ 2^7\cdot 7,2^7\cdot 3,2^7\cdot 3$ $ 2^7\cdot 10,2^7\cdot 10,2^7\cdot 3$ $ 2^8\cdot 5,2^8\cdot 5,2^8\cdot 3$ $ 2^8\cdot 5,2^8\cdot 8,2^8\cdot 8$ $ 2^{11}\cdot 5,2^{11}\cdot 1,2^{11}\cdot 1$ $ 2^{11}\cdot 6,2^{11}\cdot 6,2^{11}\cdot 1$ $ 2^{12}\cdot 3,2^{12}\cdot 3,2^{12}\cdot 1$ $ 2^{12}\cdot 3,2^{12}\cdot 4,2^{12}\cdot 4$ $ 2^{14}\cdot 3,2^{14}\cdot 1,2^{14}\cdot 1$ $ 2^{14}\cdot 4,2^{14}\cdot 4,2^{14}\cdot 1$ $ 2^{16}\cdot 1,2^{16}\cdot 1,2^{16}\cdot 1$ (Note that it might have terminated before reaching all powers of $ 2$, but that doesn't matter)

If $ n$ is odd, then start with the numbers $ (2,2,1,\dots,1)$. Because we always have either a new maximal value which occurs twice or the total number of maximal values increases by two, there is always an even amount of maximal values, so there can't be $ n$ equal numbers ($ n$ is odd). If $ n$ is even, group the numbers in pairs like Akashnil did. Now we define the permutation $ \sigma$ as follows: If $ a_i$ is even, then $ \sigma_i = i$, and if $ a_i$ and $ a_j$ are both odd, then $ a_i > a_j \iff a_{\sigma_i} < a_{\sigma_j}$ (that is, the odd numbers are sorted in the opposite way). Note that $ \sigma\circ\sigma$ is the identity, so all cycles have length of no more than 2. Now, if $ a_i$ is odd, then replace it and $ a_{\sigma_i}$ by $ a_i+a_{\sigma_i}$ (obviously, this is done only once for each pair). Then all numbers are even, and dividing by 2 won't change the result in any way. Now set $ f(a) = \sum_{i=1}^m a_i^2$ and identify the algorithm with a function $ \phi$. Then $ f(a)-f(\phi(a)) = \sum_{i=1}^m a_i^2 - \left(\sum_{a_i\text{ even}} \frac{a_i^2}{4}+\sum_{a_i\text{ odd}} \left(\frac{a_i-a_{\sigma_i}}{2}\right)^2\right) = \frac{3}{4}\sum_{a_i\text{ even}} a_i^2 + \sum_{a_i\text{ odd}} \left(\frac{a_i-a_{\sigma_i}}{2}\right)^2 \ge 0$ shows that iteration of $ \phi$ will make $ f$ decrease. However, $ f$ must be nonnegative; so $ f$ must be constant from some point on, and this is only the case if there are no even numbers and $ a_{\sigma_i} = a_i$, but by the definition of $ \sigma$, this means that minimum and maximum must be equal, and thus that all numbers must be equal.