Answer: 407

Small Hint:

Every prime dividing $n$ must divide $m,$ hence divides $m + n$ — so all prime factors of $n$ are at least $11.$

Big Hint:

Try $n = 11^2$ and $m = 11t$ with $t$ not divisible by $11;$ then $n^n$ dividing $m^m$ needs $m \ge 242.$ Scan upward for $\gcd(m + n, 210) = 1,$ then use the resulting upper bound to rule out larger deficient exponents and larger $n.$

Solution:

If a prime $p$ divides $n,$ then $p$ divides $n^n,$ which in turn divides $m^m,$ so $p$ divides $m$ and hence $p$ divides $m + n.$ 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 $p.$ Since $n^n$ divides $m^m,$ comparing exponents of $p$ gives $bm \ge an,$ so $m \ge \frac{a}{b}n.$ In particular $a \ge 2,$ so $p^2$ divides $n$ 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 $n^n$ divides $m^m,$ and $m + n = 407 = 11 \cdot 37$ is coprime to $210.$

It remains to prove that nothing smaller works. Suppose $m + n \lt 407.$ Then $n \lt 407.$ For the deficient prime above, $a \ge 3$ would imply $n \ge 11^3 \gt 407,$ so $a = 2$ and $b = 1.$ Thus $m \ge 2n,$ and $m + n \lt 407$ implies $n \le 135.$ The only integer in this range divisible by the square of a prime at least $11$, with no prime factors below $11,$ is $n = 121.$ The preceding check exhausts every possible $m$ for this $n$ below $286,$ so the least possible value is $407.$