Problem 1813

1. **Define the function $f(a,b)$:**
$$f(a,b) = a \lceil b\nu \rceil - b \lfloor a\nu \rfloor$$
We need to show that the number of excellent pairs $(a,b)$ is equal to the sum of the positive divisors of $m$.

2. **Relate $f(a,b)$ to $f(a+b,b)$ and $f(a,b+a)$:**
- If $\{a\nu\} + \{b\nu\} \geq 1$, then:
$$f(a+b,b) = (a+b) \lceil b\nu \rceil - b \lfloor (a+b)\nu \rfloor = a \lceil b\nu \rceil - b \lfloor a\nu \rfloor + b (\lceil b\nu \rceil - \lfloor b\nu \rfloor) = f(a,b) + b$$
because $\lfloor (a+b)\nu \rfloor = \lfloor a\nu \rfloor + \lfloor b\nu \rfloor$.

- Similarly, for $f(a,b+a)$:
$$f(a,b+a) = a \lceil (b+a)\nu \rceil - (b+a) \lfloor a\nu \rfloor = a (\lceil b\nu \rceil + \lceil a\nu \rceil - 1) - (b+a) \lfloor a\nu \rfloor = a \lceil b\nu \rceil - b \lfloor a\nu \rfloor = f(a,b)$$

- When $\{a\nu\} + \{b\nu\} < 1$, we have:
$$f(a,b) = f(a+b,b) = f(a,b+a) - a$$

3. **If $(a,b)$ is excellent, then $\min(a,b) \leq m$:**
- For $c,d \geq 0$, $f(c,d) = c \lceil d\nu \rceil - d \lfloor c\nu \rfloor \geq 0$.
- If $(a,b)$ is excellent, then $f(a-b,b)$ and $f(a,b-a)$ are not equal to $f(a,b)$.
- If $\min(a,b) = a$, then:
$$0 \leq f(a,b-a) = f(a,b) - a = m - a \implies a \leq m$$
- Similarly, if $\min(a,b) = b$, then:
$$0 \leq f(a-b,b) = f(a,b) - b = m - b \implies b \leq m$$

4. Define relevant pairs and branches:
- A pair $(a,b)$ is relevant if $a, b$ are positive integers, $\min(a,b) \leq m$, and $f(a,b) \leq m$.
- Draw arrows from $(a,b) \rightarrow (a+b,b)$ and $(a,b) \rightarrow (a,b+a)$.
- This forms a tree due to the Euclidean Algorithm's uniqueness.

5. Prove that relevant branches are finite:
- Suppose $B_{(a,b)}$ is infinite. Choose an infinite path of pairs $(c_i,d_i)$ for $i = 1, 2, \ldots$.
- Since $\min(c_i,d_i) \leq m$, there exists $M$ such that $c_i = c$ for all $i \geq M$.
- $f(c_i,d_i) \leq m$ and is non-decreasing, so it is eventually constant.
- This leads to a contradiction because $\{c\nu\} + \{(d_M + (i-M)c_i)\nu\} \geq 1$ for all $i \geq M$, which is impossible for an irrational $\nu$.

6. Count solutions and excellent pairs:
- Let $N(a,b)$ be the number of solutions for $x, y \in \mathbb{Z}_{\geq 0}$ in $ax + by + f(a,b) = m$.
- Let $g(a,b)$ be the number of excellent pairs in $B_{(a,b)}$.
- We claim $N(a,b) = g(a,b)$.

7. **Lemma: $N(a,b) = N(a+b,b) + N(a,b+a)$:**
- If $\{a\nu\} + \{b\nu\} \geq 1$:
$$(a+b)x + by + f(a+b,b) = m \iff ax + b(x+y+1) + f(a,b) = m \iff ax' + by' + f(a,b) = m$$
where $x' = x$ and $y' = x + y + 1$.
- Similarly:
$$ax + (b+a)y + f(a,b+a) = m \iff a(x+y) + by + f(a,b) = m \iff ax' + by' + f(a,b) = m$$
where $x' = x + y$ and $y' = y$.

8. **Prove $g(a,b) = N(a,b)$:**
- If $(a,b)$ is excellent, then $g(a,b) = 1$ and $N(a,b) = 1$ because $x = y = 0$.
- Otherwise, $g(a,b) = g(a+b,b) + g(a,b+a)$ because the set of arrows and pairs form a tree.
- Using induction, $g(a,b) = N(a,b)$ for all relevant $(a,b)$.

9. Finishing the proof:
- Let $k(m)$ be the number of excellent pairs for $m$.
- Partition the set of positive integer pairs $(a,b)$ into sets $T_1, T_2, \ldots$ where $(a,b) \in T_{\gcd(a,b)}$.
- Then:
$$k(m) = \sum_{i=1}^\infty g(i,i) = \sum_{i=1}^\infty N(i,i)$$
- We know:
$$N(i,i) = \begin{cases} 0 & \text{if } i \nmid m \\ \frac{m}{i} & \text{if } i \mid m \end{cases}$$
- Therefore:
$$k(m) = \sum_{i=1}^\infty N(i,i) = \sum_{d \mid m} N(d,d) = \sum_{d \mid m} \frac{m}{d} = \sum_{d \mid m} d = \sigma(m)$$

$\blacksquare$

The final answer is $\boxed{ \sigma(m) }$