Answer: 407

Solution:

If a prime $p$ divides $n,$ then $p \mid n^n \mid m^m,$ so $p \mid m$ and hence $p \mid m + n.$ Since $\gcd(m + n, 210) = 1,$ no prime factor of $n$ is $2,$ $3,$ $5,$ or $7:$ every prime factor of $n$ is at least $11.$ Because $m$ is not a multiple of $n,$ some prime $p$ has $b = v_p(m) \lt a = v_p(n),$ where $v_p$ denotes the exponent of $p.$ Comparing exponents of $p$ in $n^n \mid m^m$ gives $bm \ge an,$ so $m \ge \frac{a}{b}n \ge 2n.$ In particular $a \ge 2,$ so $p^2 \mid n$ and $n \ge 11^2 = 121.$

Take $n = 121$ with $p = 11,$ $a = 2,$ $b = 1:$ then $m$ is a multiple of $11$ but not of $121,$ and $m \ge 2 \cdot 121 = 242.$ The candidates $m = 253, 264, 275$ give $m + n = 374 = 2 \cdot 11 \cdot 17,$ $385 = 5 \cdot 7 \cdot 11,$ $396 = 2^2 \cdot 3^2 \cdot 11,$ all sharing a factor with $210,$ while $m = 242 = 2 \cdot 11^2$ is a multiple of $121.$ But $m = 286 = 2 \cdot 11 \cdot 13$ works: $v_{11}(m^m) = 286$ $\ge 242 = v_{11}(n^n),$ so $n^n \mid m^m,$ and $m + n = 407 = 11 \cdot 37$ is coprime to $210.$

Any other admissible $n$ is at least $13^2 = 169,$ forcing $m + n \ge 3n \ge 507.$ Hence the least possible value is $407.$