Consider the collection $$M$$ of all ordered pairs $$(a,b)$$ of positive integers $$a$$ and $$b$$ which satisfy 

$$ ab=406+11\cdot\operatorname{lcm}(a,b)+7\cdot\gcd(a,b).$$ 

What is the smallest possible value of $$a+b$$?

Let $$g=\gcd(a,b)$$. Write $$a=gx,\;b=gy$$ where $$x,y\in\mathbb{N}$$ and $$\gcd(x,y)=1$$.

Then
$$ab=g^2xy,$$
$$\operatorname{lcm}(a,b)=gxy.$$

The given relation becomes
$$g^2xy \;=\;406+11(gxy)+7g.$$

Bring every term to the left and factor out $$g$$:
$$g^2xy-11gxy-7g-406=0$$
$$\Longrightarrow\;g\bigl[xy(g-11)-7\bigr]=406.$$

Set
$$k=xy(g-11)-7.$$
Then $$gk=406.$$ Because $$a,b$$ are positive, $$xy\gt0$$, so $$k\gt0$$ and consequently $$g-11\gt0\;(\text{i.e. }g\gt11).$$

All positive divisors of $$406=2\cdot7\cdot29$$ are $$1,2,7,14,29,58,203,406.$$ The divisors exceeding $$11$$ are $$14,29,58,203,406.$$

Then $$k=\dfrac{406}{14}=29,$$ and

$$xy(g-11)=xy\cdot3=k+7=36\;\Longrightarrow\;xy=12.$$

Since $$x,y$$ are coprime and $$xy=12$$, the admissible pairs are $$(1,12),\,(3,4),\,(4,3),\,(12,1).$$ For each pair $$a+b=g(x+y)=14(x+y).$$ The smallest sum arises from $$(x,y)=(3,4)\text{ or }(4,3):$$ $$a+b=14(3+4)=98.$$

Now $$k=\dfrac{406}{29}=14$$ and $$xy(g-11)=xy\cdot18=21\;\bigl(k+7\bigr),$$ giving $$xy=\dfrac{21}{18}$$ - not an integer. Hence no solution.

Then $$k=7$$ and $$xy(g-11)=xy\cdot47=14,$$ which again yields a non-integer $$xy$$. No solution.

Here $$k=2$$ and $$xy(g-11)=xy\cdot192=9,$$ impossible.

Here $$k=1$$ and $$xy(g-11)=xy\cdot395=8,$$ impossible.

The only feasible case is $$g=14$$ with $$(x,y)=(3,4)\text{ or }(4,3).$$ Taking $$a=42,\;b=56$$ (or vice-versa) indeed satisfies

$$ab=42\cdot56=2352,$$ $$406+11\cdot\operatorname{lcm}(42,56)+7\cdot\gcd(42,56)=406+11\cdot168+7\cdot14=2352.$$

Therefore the minimum possible value of $$a+b$$ is $$\mathbf{98}$$.