1. Define the function and set up the problem:
Let $\sigma(n)$ be the sum of the divisors of $n$ and define $f(n) = \dfrac{\sigma(n)}{n}$. We are interested in finding a set $M$ such that for all $a, b \in M$, $f(a+b) < 2$.

2. Construct the sequence of sets:
Start with $M_1 = \{1\}$. We will construct a sequence of sets $M_1, M_2, \ldots$ such that $M_{k+1} = M_k \cup \{a_{k+1}\}$ for some positive integer $a_{k+1}$ with $a_1 = 1$. The union of all these sets will produce an infinite set $M$.

3. Choose the elements of the sequence:
Let $p_1 < p_2 < \ldots$ be the primes in increasing order. Suppose at some step we have the set $M_k = \{a_1, a_2, \ldots, a_k\}$. We will choose $a_{k+1} = p_1 p_2 \ldots p_m + 1$ for sufficiently large $m$ to be determined later. We need $a_i + a_{k+1}$ to be a good number for each $i \le k+1$.

4. Lemma:
For all $n$, $f(n) < \prod_{p|n} \left( \dfrac{p}{p-1} \right)$.
- Proof:
Note that $f(n) = \prod_{p|n} \left( 1 + p^{-1} + \ldots + p^{-v_p(n)} \right) < \prod_{p|n} \left( \dfrac{p}{p-1} \right)$ by the infinite geometric series formula applied to each term of the product.

5. **Show that $2a_{k+1}$ is a good number:**
For large $m$, our choice of $a_{k+1}$ satisfies that $2a_{k+1}$ is a good number. All primes dividing $a_{k+1}$ must have size at least $p_{m+1}$, hence:
$$f(2a_{k+1}) = 1.5 f(a_{k+1}) < 1.5 \prod_{p|a_{k+1}} \left( \dfrac{p}{p-1} \right) \le 1.5 \prod_{p|a_{k+1}} \left( \dfrac{p_{m+1}}{p_{m+1}-1} \right)$$
Since $a_{k+1} = p_1 p_2 \ldots p_m + 1 \le p_{m+1}^m$, this term is at most:
$$1.5 \cdot \left( 1 + \dfrac{1}{p_{m+1}-1} \right)^m$$
For large enough $m$, the ratio $b_m = \dfrac{m}{p_{m+1}-1}$ grows arbitrarily small (as roughly $(\log m)^{-1}$), hence:
$$1.5 \cdot \left( 1 + \dfrac{1}{p_{m+1}-1} \right)^m \approx 1.5 \cdot e^{b_m} < 2$$

6. **Show that $a_i + a_{k+1}$ is a good number:**
For sufficiently large $m$, each $a_i + a_{k+1}$ must be a good number. Assume $p_m > a_k$. Then:
$$a_i + a_{k+1} = (a_i + 1) + p_1 p_2 \ldots p_m$$
By the inductive hypothesis, $a_i + 1 = a_i + a_1$ is good, hence $f(a_i + 1) = c_i < 2$. Since $p_m \ge a_k + 1 \ge a_i + 1$, if $p \nmid a_i + 1$ and $p | a_i + a_{k+1}$, we know $p \nmid p_1 p_2 \ldots p_m \implies p \ge p_{m+1}$. It follows that:
$$f(a_i + a_{k+1}) < \prod_{p|a_i + a_{k+1}, p|a_i + 1} \left( \dfrac{p}{p-1} \right) \prod_{p|a_i + a_{k+1}, p \ge p_{m+1}} \left( \dfrac{p}{p-1} \right) \le c_i \prod_{p|a_i + a_{k+1}, p \ge p_{m+1}} \left( \dfrac{p_{m+1}}{p_{m+1}-1} \right)$$
For each $a_i + a_{k+1}$ to be good, we need:
$$\left( \dfrac{p_{m+1}}{p_{m+1}-1} \right)^{\log_{p_{m+1}} (a_i + a_{k+1})} \le \dfrac{2}{c_i}$$
where each $c_i$ is less than $2$. Since $a_i + a_{k+1} < 2(p_1 p_2 \ldots p_m + 1) < 4 p_2 p_3 \ldots p_m \le p_{m+1}^m$, the LHS of the previous inequality is at most:
$$\left( \dfrac{p_{m+1}}{p_{m+1} - 1} \right)^m$$
As with the previous argument, for large $m$ this term approaches $e^{b_m}$, which grows arbitrarily close to one as $b_m \to 0$. Hence for suitable $m$ we have:
$$\text{LHS} \le \min \left( \dfrac{2}{c_1}, \ldots, \dfrac{2}{c_k} \right)$$
Therefore, each $a_i + a_{k+1}$ is good, as desired.