Given two coprime integers $a < b$ of different parities, only a finite number of points in $\mathbb N^2$ cannot be reached by a walk in $\mathbb N^2$, starting at the origin and using only steps of the form $(b,\pm a),(\pm a,b)$ (and thus making an acute angle with the north-eastern vector $(1,1)$).
Is there a good upper bound on the number of such exceptional points? Is there a good upper bound on the coordinate sum $x+y$ of such an exceptional point $(x,y)$?
(Remark: A naive proof that almost all points can be reached gives an upper bound which is probably very far from the true value.)
Added in order to comply with the request made by user 9072:
Proof that almost all points can be reached: We consider $v=a(b,a)+b(-a,b)=(0,a^2+b^2)$ and $w=(a,b)+(-a,b)=(0,2b)$.
Since $a,b$ are coprime and have different parities, $a^2+b^2$ and $2b$ are coprime. The set $$\{k(a^2+b^2),k=0,\ldots,2b-1\}$$ with largest element $(2b-1)(a^2+b^2)<4b^3$ contains thus representants of all classes of $\mathbb Z/(2b)\mathbb Z$. This shows that all elements $(0,y)$ with $y\geq 4b^3$ are of the form $\mathbb N v+\mathbb N w$. Using the obvious symmetry with respect to coordinates, the same holds for $(x,0)$ with $x\geq 4b^3$. We get thus everything in $4b^3(1,1)+\mathbb N^2$. All non-reachable points are thus contained in the two strips of width $4b^3-1$ parallel to coordinate axes. Let $(x,y)$ with $x=\min(x,y)<4b^3$ be a point of the vertical strip. If $y\geq \left\lceil\frac{4b^3}{a}\right\rceil b+4b^3$, then $$(x,y)\in \mathbb N(-a,b)+(4b^3)(1,1)+\mathbb N^2$$ and we are done.
This gives the very crude upper bound $64b^7$ on the number of non-reachable points in $\mathbb N^2$.
Some computational data:
For $a=1$ and $b\in\{2,4,6,\ldots,16\}$ the number of missing elements is given by $$\frac{1}{12}(4b^5-3b^4+8b^3)$$ and the largest coordinate occuring in missing points is given by $b^3-b$.
For $a=2$ and $b\in\{3,5,7,9,\ldots,17\}$ the number of missing elements is given by $$\frac{1}{12}(2b^5+3b^4-2b^3-24b^2+12b+9)$$ and the largest coordinate occuring in missing points is given by $$\frac{1}{2}(b^3+2b^2-9b)$$ if $b>3$ and by $10$ for $b=3$ (the polynomial formula yields the wrong value $9$).
For $a=3$ and $b\in\{4,8,10,14,16,20,22,26\}$ the number of missing elements is given by $$\begin{cases}\frac{1}{36}(4b^5+15b^4-8b^3-240b^2+160b+384)\qquad&b\equiv 1\pmod 3,\\ \frac{1}{36}(4b^5+15b^4-8b^3-240b^2+184b+360)&b\equiv 2\pmod 3\end{cases}$$ and the largest coordinate occuring in missing points is given by $$\begin{cases}\frac{1}{3}(b^3+4b^2-17b)\qquad&b\equiv 1\pmod 3,\\ \frac{1}{3}(b^3+4b^2-15b)&b\equiv 2\pmod 3.\end{cases}$$
For $a=4$ and $b\in\{5,7,9,\ldots,21\}$ the number of missing elements is given by $$\begin{cases}\frac{1}{12}(b^5+6b^4-b^3-162b^2+120b+540)\qquad&b\equiv 1\pmod 4,\\ \frac{1}{12}(b^5+6b^4-b^3-162b^2+132b+528)&b\equiv 3\pmod 4\end{cases}$$ and the largest coordinate occuring in missing points is given by $$\frac{1}{4}(b^3+6b^2-27b).$$
There seems to be some structure in these data: Fixing $a$, in all examples the number of missing elements seems to be of the form $$\frac{1}{3a}b^5+\frac{3a-4}{4a}b^4+O(b^3)$$ and the largest coordinate occuring in missing elements seems to be of the form $$\frac{1}{a}b^3+\frac{2(a-1)}{a}b^2+O(b)\ .$$
