Time for some induction

Combinatorial proof:

Consider the set $S$ of points $(x,y)$ on the plane where $1 \le y \le x \le n$ and $x,y$ are positive integers.

Consider a line $y = \frac{m}{k} x$ where $1 \le m \le k \le n$ and $m,k$ are coprime. This line passes exactly $t = \left\lfloor \frac{n}{k} \right\rfloor$ points of $S$ , namely $(k,m), (2k,2m), (3k,3m), \ldots, (tk,tm)$ . ( $x$ has to be a positive integer in order for the point to be an element of $S$ , and moreover a multiple of $k$ so that $y$ is an integer, but any such positive integer not greater than $n$ works and gives a unique point for $S$ .) Let $S_k$ be the set of all points of $S$ lying on a line in the form $y = \frac{m}{k} x$ where $1 \le m \le k$ , $m,k$ are coprime. There are $\phi(k)$ choices of $m$ by definition of $\phi$ , and each of these gives $\left\lfloor \frac{n}{k} \right\rfloor$ points, so $S_k$ contains $\left\lfloor \frac{n}{k} \right\rfloor \phi(k)$ points.

Now, consider $S_1, S_2, \ldots, S_n$ . We will show that every point of $S$ is in at least one of the sets. Consider a point $(x_0, y_0) \in S$ . This point lies on the line $y = \frac{y_0}{x_0} x$ where $1 \le y_0 \le x_0 \le n$ . Suppose $x_0 = cx_1, y_0 = cy_1$ , where $x_1, y_1$ are coprime; then simplifying the fraction gives that $(x_0, y_0)$ lies on the line $y = \frac{y_1}{x_1} x$ . By definition of $S_i$ , this means $(x_0, y_0)$ is in $S_{x_1}$ . Thus every point of $S$ is in at least one of the sets.

We will also show that, on the other hand, every point of $S$ is in at most one of the sets. Consider a point $(x_0, y_0) \in S$ . Suppose for the sake of contradiction that this point is in two sets $S_i, S_j$ ; without loss of generality, suppose $i < j$ . Then there exists $m_i, m_j$ such that $y_0 = \frac{m_i}{i} x_0$ and $y_0 = \frac{m_j}{j} x_0$ , where $m_i, i$ are coprime, and $m_j, j$ are coprime. This implies $\frac{y_0}{x_0} = \frac{m_i}{i} = \frac{m_j}{j}$ . But a fraction only has one reduced form (with a positive denominator), while if $m_i, i$ are coprime and $m_j, j$ are coprime, both of $\frac{m_i}{i}, \frac{m_j}{j}$ are reduced fractions, contradiction. Thus no point of $S$ is in more than one of the sets.

In total, this means each point of $S$ is in exactly one of the sets $S_1, S_2, \ldots, S_n$ . Thus $\sum |S_i| = |S| = 1+2+\ldots+n = \frac{n(n+1)}{2}$ . But $|S_i| = \left\lfloor \frac{n}{k} \right\rfloor \phi(k)$ , so $\sum |S_i|$ is the given expression. Thus $A = 1, B = 2$ , and $A+B = \boxed{3}$ .

We shall prove that the statement is true for n=1: $\sum_{k=1}^1\lfloor{\frac{1}{k}}\rfloor \phi(k)=1\times \phi(1)=1=\frac{1(1+1)}{2}$ , which is indeed true.

Taking the inductive step, we now assume that the statement is true up to a number n, and then we will prove it true for the number n+1. We observe that: $\sum_{k=1}^n\lfloor{\frac{n}{k}}\rfloor \phi(k)=\sum_{k=1}^\infty\lfloor{\frac{n}{k}}\rfloor \phi(k)$ because $\lfloor{\frac{n}{k}}\rfloor=0$ for every $k\geq{n+1}$

Then using the fact that every number n for any given k can be written in the form: $n=kq(n,k)+r(n,k),\ 0\leq r(n,k)<k$ , we can set out to prove that: $\lfloor\frac{n}{k}\rfloor=\lfloor\frac{kq+r}{k}\rfloor=\lfloor q+\frac{r}{k}\rfloor=q(n,k)$ , since $\frac{r}{k}<1$ and q is an integer.

In the light of the above, the statement for the number n+1 reads : $\sum_{k=1}^\infty\lfloor{\frac{n+1}{k}}\rfloor \phi(k)=\sum_{k=1}^\infty q(n+1,k)\phi(k)=\sum_{k=1}^\infty (q(n,k)+\lfloor \frac{r(n,k)+1}{k}\rfloor)\phi(k)$

However, we note that : $\lfloor \frac{r(n,k)+1}{k}\rfloor=\begin{cases}1 & \quad \text{if } n+1\equiv 0\ mod \ {k}\\ 0 & \quad \text{all other cases} \\ \end{cases}$

We conclude that the second part of sum contributes only whenever k|n+1 and thus : $\sum_{k=1}^\infty\lfloor{\frac{n+1}{k}}\rfloor \phi(k)=\sum_{k=1}^\infty\lfloor{\frac{n}{k}}\rfloor \phi(k)+\sum_{k|n+1}\phi(k)=\frac{n(n+1)}{2}+n+1=\frac{(n+1)(n+2)}{2}$ , which concludes the proof.

In order to prove that $\sum_{k|n}\phi(k)=n$ we first calculate for p a prime number: $\sum_{p^{a}|n}\phi(k)=\sum_{k=0}^{a-1}\phi(p^k) =(1-\frac{1}{p}) \sum_{k=1}^{a-1}p^k+1=(1-\frac{1}{p})\frac{p}{p-1}(p^{a}-1) +1=p^a$

and then we assert that: $\sum_{k|p_{1}^{a_{1}}...p_{l}^{a_{l}}}\phi(k)=\sum_{k|p_{1}^{a_{1}}\hspace{0.1cm} or ... or \hspace{0.1 cm} k| p_{l}^{a_{l}}}\phi(k)=\prod_{i=1}^{l}\sum_{p_{i}^{a_{i}}|n}\phi(k)=\prod_{i=1}^{l}p_{i}^{a_{i}}$ which means that for generic $n=\prod_{i=1}^{l}p_{i}^{a_{i}}$ , the lemma holds.

Sorry, this is not a proper solution, but I think this is quite interesting guess.

Notice that $\frac{n(n+A)}{B}$ must have an integer solution. Since $n$ is an arbitrary integer, then $B$ has to be either $1$ or $2$ ; otherwise, $\frac{n(n+A)}{B}$ is not integer. Next, we will show that $B$ can't be $1$ plug-in $n=1$ and $n=2$ for example. Thus, $B=1$ . $A$ can be easily obtained afterwards.