$gcd(2a-1,2b+1)=1$ => $gcd(2a-1,a+b)=1$ and $gcd(2b+1,a+b)=1$ $a+b|(4ab+1)(a-b+1)+2[(a+b)^{2}-4ab-1]=(a-b+1)(2a-1)(2b+1)$ then $a+b|a-b+1$ => $(a,b)\in \{(0,1),(1,0)\}$

$(a;b)=(n;n+1)$

$a+b | 4ab+1$ $\longrightarrow$ $a+b | 4a^{2}-1$ $\longrightarrow$ $a+b|(2a-1)(2a+1)$ , $a+b|(2b-1)(2b+1)$ , if there exist $p$ such that $p|2a-1$ and $p|a+b$ then $p|(2b-1)(2b+1)$ , but $2a-1$ and $2b+1$ are coprime $\longrightarrow$ $p|2b-1$ $\longrightarrow$ $p|2a-2b$,$p|2a+2b$ $\longrightarrow$ $p|4a$ but $p|2a-1$ . contradiction . now we have $(a+b,2a-1)=1$ and we have $(a+b,2b+1)=1$ $\longrightarrow$ $a+b|2a+1$ , $a+b|2b-1$ $\longrightarrow$ $2a+1=m(a+b)$ and $2b-1=n(a+b)$ and $m,n$ are natural numbers because $a$ and $b$ are natural . $\longrightarrow$ $(2a+1)+(2b-1)=(m+n)(a+b)$ $\longrightarrow$ $m+n=2$ $\longrightarrow$ $m=n=1$ $\longrightarrow$ $b=a+1$ $(a;b)=(n;n+1)$

The answer $(a,b)=(n,n+1)$ for every integer $n$. It is clear from the fact $a+b|4ab+1$ that $a+b|(2a+1)(2b+1)$ and $a+b|(2a-1)(2b-1)$ Now let $gcd(2a-1,a+b)=g$.Hence $g|2(a+b)$ and $g|2a-1$ implies that $g|2b+1$ which contradicts our assumption that $gcd(2a-1,2b+1)=1$.So $gcd(a+b,2a-1)$.So $a+b|2b-1$.But $2b+1<2(a+b)$.Hence $2B+1=2a+2b$.And thus $b=a+1.$.Now it is clear that then $a+b|4ab+1$.Again $gcd(2a-1,2b+1)=gcd(2b+1,4)=1$.Thus every pair of this form satisfies the condition. So answer is $(a,b)=(n,n+1)$ $\forall$ $n$.