Start with the definitions of $\sigma$ and $\phi$:
$ \sigma(n) = \sum_{d \vert n}{d}$
$ \phi(n) = \prod_i{(p_i - 1) \cdot p_i^{e_i-1}} $
Expand the product and rewrite as:
$ \phi(n) = \sum_{d \vert n}{a_d \cdot d}$
with coefficients $a_d \in \{-1,0,1\}$. So now we have
$ \sigma(n) + \phi(n) = \sum_{d \vert n}{(a_d + 1) \cdot d} $
and since $a_n = 1$:
$ \sigma(n) + \phi(n) = 2 \cdot n + \sum_{d \vert n, d < n}{(a_d + 1) \cdot d}$
Now $\sigma(n) + \phi(n) = 2 \cdot n$ if and only if $a_d = -1$ for all proper divisors $d$. This is only true for prime $n$, and $n=1$. Otherwise some positive terms remain.
Yes, every prime occurs as a prime factor of a highly composite number. Suppose towards a contradiction that it does not; then let $p_n$ be the least prime which does not occur as a factor. Note that this means a highly composite number cannot contain any primes higher than $p_n$ either, since switching out such a prime for $p_n$ gives a smaller number with the same number of divisors. Thus, this would mean that all highly composite numbers are of the form
$$
2^{k_1} \cdot 3^{k_2} \cdots p_{n-1}^{k_{n-1}}.
$$
Since there are infinitely many highly composite numbers, at least one of the values $k_i$ must be unbounded; hence, there is a prime $p_i$ and a highly composite number such that $p_n < p_i^{\lfloor k_i/3 \rfloor }$. But now note that removing $\lfloor k_i/3\rfloor$ copies of the prime $p_i$ from this number lowers the total number of divisors by less than $1/3$, while adding the prime $p_n$ doubles its number of divisors. This contradicts the assumption that this was the prime factorization of a highly composite number.
Thus, $p_n$ occurs in the prime factorization of a highly composite number.
Edit: Since this question has received so much attention, I feel like it might be worth explaining some of the intuition behind this answer. Recall, a highly composite number is a number which has more divisors than any number before it. Now, if
$$
p_1^{k_1} \cdots p_n^{k_n}
$$
is any prime factorization of a number, then that number will have
\begin{equation}
\qquad\qquad\qquad\qquad\qquad(1 + k_1) \cdots (1 + k_n)\qquad\qquad\qquad\qquad(1)
\end{equation}
divisors: namely, for every prime $p_i$ a divisor can include that prime $0, 1, \ldots, k_i$ times, which gives $1 + k_i$ options. Note that the primes don't really factor into this anymore; swapping out all copies of a prime with a prime that did not yet occur in the number leaves the $k_i$ and therefore the number of divisors unchanged.
Of course, which primes occur in the number does matter for the size of the number: higher primes give a higher number.
Thus, we think of constructing highly composite numbers as a sort of optimization problem: by multiplying by a new prime factor, we "buy" some new divisors, at the "expense" of making the number larger.
At first, we use low prime factors for that; multiplying by 2, 3, 5 is much "cheaper" than multiplying by 9999991. However, there is a law of diminishing returns. The first time you add a prime factor, you double the number of divisors: in formula (1), a factor $(1 + 0)$ is replaced by one of the form $(1 + 1)$. The second time you do it, a factor $(1 + 1)$ is replaced by $(1 + 2)$, providing only a multiplier of $\frac32$ to the number of divisors.
So, the more and more of the same prime factor you add, the less returns you get off that. If you keep it up long enough, higher primes become more "lucrative". Eventually, you will have exhausted so much of the usefulness of all primes below 9999991, that it becomes optimal to add the prime factor 9999991. The proof just makes this precise. With a little more fidgeting, you can prove that in fact every number, not just every prime, divides a highly composite number.
Best Answer
Consider a given positive integer $n \ge 1$, and a corresponding positive integer $K$, for which a prime $P$ exists such that $P$, $P + K$, $P + 2K$, $\ldots$, $P + nK$ are all primes. Note $K$ must be an integral multiple of the primorial $p_j\#$, where the index refers to the prime index (e.g., $p_1 = 2$, $p_2 = 3$, etc.) and $p_j$ is the largest prime $\le n$. Thus, the smallest possible $K$ would be $p_j\#$ itself. This is why the minimum $K$ for your example where $n = 5 = p_3$ is $30$ since $30 = p_3\# = 2(3)(5)$.
To prove $p_j\# \mid K$, first note if $P = p_i$ for some $1 \le i \le j$, then $p_i \mid P + p_i(K)$, so it can't be prime, which means $P \ge p_{j+1}$. Next, assume there's a $p_i$, with $i \le j$, where $p_i \nmid K$. Then $P$, $P + K$, $\ldots$, $P + (p_i - 1)K$ all have different remainders modulo $p_i$ (since if any $2$ were the same, say $P + qK$ and $P + rK$ with $r \gt q$, the difference of $(r - q)K$ must be divisible by $p_i$ but $0 \lt r - q \lt p_i$). Since there are $p_i$ possible remainders from $0$ to $p_i - 1$ among these $p_i$ values, one of them must be $0$. As it must be $\gt p_i$, it can't be prime. This shows the original assumption that $p_i \nmid K$ must be false.
A special case to consider is if $n + 1$ is prime, i.e., it's $p_{j+1}$. If so, then if $p_{j+1} \nmid K$, you must have $P = p_{j+1}$ since, otherwise, one of the $P + iK$ for $1 \le i \le n$ must be a multiple of $p_{j + 1}$ and $\gt p_{j+1}$, so it can't be prime.
Regarding proving there always exists a prime $P$ where, with $K = p_j\#$, you have $P + ik \; \forall \; 1 \le i \le n$ being prime, there's no general method I know of to prove this. Although I suspect such a $P$ always exists, the only thing known for sure now is that a $K$ and $P$ do exist for any positive $n$, as explained in the next paragraph.
As for finding the largest $n$ where there's a $K$, including a smallest such $K$, where $P$, $P + K$, $\ldots$, $P + nK$ are all prime, there is no such maximum $n$. Note the Green–Tao theorem states