We have so much in common

First, we know that one of $n$ or $n + 1$ is even. Suppose $n$ is even; then the divisors of $n$ are $1, 2, \frac{n}{2}$ and $n$ . The divisors of $n + 1$ will then be $1, p, \frac{n + 1}{p}$ and $n + 1$ for some prime $p \gt 2.$ For the divisors of both numbers to have the same sum we must have

$1 + 2 + \frac{n}{2} + n = 1 + p + \frac{n + 1}{p} + (n + 1)$

$\Longrightarrow 1 + \frac{n}{2} = p + \frac{n + 1}{p}$

$\Longrightarrow \frac{n}{2} - \frac{n}{p} = p + \frac{1}{p} - 1$

$\Longrightarrow n = \dfrac{2(p^{2} - p + 1)}{p - 2} = 2(p + 1) + \dfrac{6}{p - 2}.$

We see that $n$ will be integral only for $p = 3$ and $p = 5$ , which both yield $n = 14.$

If it were $n + 1$ that were even, then the divisor sums would be such that

$1 + 2 + \frac{n + 1}{2} + (n + 1) = 1 + p + \frac{n}{p} + n$

$\Longrightarrow \frac{n}{2} - \frac{n}{p} = p - \frac{7}{2}$

$\Longrightarrow n = \dfrac{p(2p - 7)}{p - 2} = 2p - 3 - \dfrac{6}{p - 2}.$

We see that $n$ will be integral only for $p = 3$ and $p = 5$ , but this time the corresponding values for $n$ will be $-3$ and $5$ , neither of which has four divisors, so this case can be discarded.

Thus there is only one element in $S$ , namely $14$ , and so $m + b = 1 + 14 = \boxed{15}.$

If a number has four divisors, it is of the form $n = pq$ with $p\not=q$ prime or $n = p^3$ with $p$ prime. We can treat these cases together by writing $n = pq\ \ \ \ \ \sigma(n) = (p+1)(q+1)\ \ \ \ \ p\ \text{prime and}\ q = p^2\ \text{or}\ q\ \text{prime}.$ Now $n+1$ must be of the same form. It is also clear that either $n$ or $n+1$ is even, and therefore has 2 as a prime factor. We take both possibilities together by transforming $(n,n+1)\to(n-1,n)$ . $n\pm 1 = 3r\ \ \ \ \ \ \sigma(n\pm1) = (2+1)(r+1)\ \ \ \ \ \ r\ \text{prime or}\ r = 2^2 = 4.$ The sums of divisors are equal: $(p+1)(q+1) = 3(r+1); \ \ \ \ \ \ \ \ pq + p + q = 3r + 2.$ Substitute $r = \tfrac12(pq\pm1)$ : $pq + p + q = \tfrac32(pq\pm1) + 2;\ \ \ \ \ \ \ \tfrac12pq - p - q + 2 \pm\tfrac32 = 0.$ We factor $\tfrac12(p - 2)(q - 2) = \mp\tfrac32;$ Since $p, q > 2$ , we choose the positive sign (corresponding to a minus sign in the original choice) and get $(p - 2)(q - 2) = 3; \ \ \ \ \ p = 3, q = 5\ \ \text{or vice versa}.$ Thus $n = 3\cdot 5 = 15;\ \ 2r = n-1 = 14, r = 7.$ $p, q, r$ are prime and therefore satisfy our conditions. We check the divisors: 15 gives $1 + 3 + 5 + 15 = 24,$ and 14 gives $1 + 2 + 7 + 14 = 24.$

Conclusion: $n = 14$ and there is only one such solution.

The number of divisors for $n=p_{1}^{a_{1}}p_{2}^{a_{2}}p_{3}^{a_{3}}...$ with $p_{m}$ prime, is given by $(a_{1}+1)(a_{2}+1)(a_{3}+1)...$ . Any number with $4$ divisors, then, must have either $a_{1}=3,a_{m}=0$ for all other $m$ or $a_{1}=1, a_{2}=1, a_{m}=0$ for all other $m$ . Meaning both $n$ and $n+1$ are of the form $p^{3}$ or $pq$ for primes $p$ and $q$ . If we take $n$ or $n+1$ to be $p^{3}$ then the other is either $p^{3}+1$ or $p^{3}-1$ which factor to have a terms either $p+1$ or $p-1$ along with another factor larger than $1$ , for primes larger than $3$ , $p+1$ and $p-1$ are both composite, meaning $p^{3}+1$ or $p^{3}-1$ has more than $4$ factors. For either $n$ or $n+1$ to be $p^{3}$ , $p$ must be $2$ or $3$ . Checking these case by case yields $(n,n+1)=(7,8)(8,9),(26,27),(27,28)$ . The only solution among these that conforms to each having $4$ factors is $(26,27)$ but the factors of these two don't sum to the same amount, hence both $n$ and $n+1$ are of the form $pq$ for primes $p$ and $q$

Since one of the two numbers must be even let the even one be $2r$ for prime $r$ and the other be $pq$ . Now we have $pq\pm1=2r$ . The sum of the factors for the even and odd one are $(2+1)(r+1)=3(r+1)$ and $(p+1)(q+1)=pq+p+q+1$ respectively and these must be equal.

Substituting $\frac{pq\pm1}{2}=r$ into $3(r+1)=pq+p+q+1$ yields $\frac{3(pq\pm1)}{2}+3=pq+p+q+1$

$\rightarrow 3(pq\pm1)+6=2pq+2p+2q+2$

$\rightarrow pq-2p-2q\pm3+4=0$

$\rightarrow (p-2)(q-2)=\mp3$

$(p-2)(q-2)$ must be positive, ruling out $pq+1=2r$ so it must be that $n=2r$ and $n+1=pq$ .

$(p-2)(q-2)=3$ gives $p=3,q=5$ and consequently $n=14$ as the only valid solution. This means $S$ has only one element, $14$ , so $m=1,b=14$ and $m+b=\boxed{15}$

14 is a number n with 4 divisors....... 1,2,7,14........=n therefore n+1=15.... it has also 4 divisors 1,3,5,15........... number of elements m in the set of numbers with four divisors=1 [s belongs to{14}]=m sum of all the elements in set S is 14=b therefore m+b=14+1=15............ Ans..... 14 is the only n that satisfies that n and n+1 both have only 4 divisors