Mr. 11 is back to the game!

By Fermat's last theorem, $x^{11} +y^{11}=z^{11}$ has no non-zero integer solutions.

Hence we conclude that at least one out of $x,y,z$ is zero.

Thus the triples $(x,y,z)$ will be of just of 4 types, for $m\neq 0$

$1. (m,0,m)\\ 2. (0,m,m) \\ 3. (m,-m,0) \\4. (0,0,0)$

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In the first three cases, $k$ has $200$ choices (-100 to 100 excluding zero), giving $200\times 3 = 600$ ordered triples. 4th case will contribute $1$ triple ,hence $n=601$

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We see that $601 \equiv 7 \equiv -4 \pmod{11}$ ,

Hence $k \equiv -1 \pmod{11}$ (Because then $601\times k \equiv (-4)(-1) \equiv 4 \pmod{11}$ )

Minimum positive integer value of $k$ is $10$ .

Answer $n+k = 601+10 =\boxed{611}$

By Fermat's last theorem, the equation has no non-zero integer solutions then we will check for 1)x=0 2)y=0 3)z=0 4)and one case in which three of them simoultaneously equal to zero