Is it a 4x4 Magic Square?

The only values are $n = 2$ and $4$ . To see this, note first that the sum of the 16 entries is $\displaystyle \sum_{k=1}^{16} k = 136$ . Let $R_i$ and $C_i$ denote the $i^{th}$ row sum and the $i^{th}$ column sum, respectively, where $i = 1, 2, 3, 4$ . Then, by assumptions, $R_i = a_in, C_i = b_in$ where the 8 natural numbers $a_i$ 's and $b_i$ 's are all distinct.

Hence $2 \times 136 = \displaystyle \sum_{i=1}^4 (R_i + C_i) = n \sum_{i=1}^4 (a_i + b_i) \geq \sum_{i=1}^8 k = 36n$ , and thus $n \leq \lfloor \frac{272}{36} \rfloor =7$ .

On the other hand, since $n | R_i$ for all $i = 1,2,3,4$ , we have $n | \displaystyle \sum_{i=1}^4 R_i$ or $n | 136$ . Since $136 = 2^3 \times 17$ , we conclude that the only possible values of $n$ are $n = 2$ or $4$ .

So sum of all possible values of $n = \boxed{6}$ .

To complete the proof it clearly suffices to display a configuration in which all the 4 row sums and the 4 column sums are distinct multiples of 4. The array shown in the figure below is one such configuration since $(R1;R2;R3;R4) = (16; 20; 48; 52)$ and $(C1; C2; C3;C4) = (28; 32; 36; 40)$ .