$m + n?$

Relevant wiki: Bezout's identity

If $1288$ leaves a remainder of $1$ when divided by $H$ this implies

$H\mid 1287$

With a similar reasoning

$H\mid 2907$

Since $H$ is by definition the biggest number which divides both $1287$ and $2907$

$H=lcm(1287,2907)=lcm(3^2\cdot 11\cdot 13,3^2\cdot 17\cdot 19)=9$

We now have

$45m+288n=9 \iff 5m+32n=1$

Using the Euclidean algorithm to find $\gcd(5,32)$ and then working backwards

$\begin{aligned} \gcd(5,32)&=\gcd(5,32-6\cdot 5)\\ &=\gcd(5,2)\\ &=\gcd(5-2\cdot 2,2)\\ &=(1,2) \end{aligned}$

$\begin{aligned} 1&=5-2\cdot 2\\ &=5-2\cdot(32-6\cdot 5)\\ &=13\cdot 5-2\cdot 32 \end{aligned}$

Hence

$m=13,n=-2\Longrightarrow m+n=\boxed{11}$

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Note:

This in fact is not completely correct, since infinitly many solutions can be generated in the following way

$13\cdot 5-2\cdot 32=1$

$13\cdot 5 +160t-2\cdot 32-160t=1$

$5(32t+13)+32(-5t-2)=1$

$\Longrightarrow m=32t+13,n=-5t-2$