The maximum number of straight making ranks Poker nerd question here... I'll start out with some definitions of standard poker terms:
A flop is $3$ cards chosen from a $52$ card deck. For example: $789$ (suits are irrelevant to this question). If you draw a fourth card from the deck you get a turn card. 
A straight is 5 cards in succession, for example $56789.$ When making a straight, an Ace can count both as $1$ and $14$. For example, $TJQKA$ is a straight, and so is $A2345$. However, you cannot "wrap around the ace": For example, $KQA23$ is not a straight.
Finally, a holdem hand is just 2 cards chosen from a 52 card deck.
Now, here's a definition I've made up myself. Let's say we are given a flop F. I define a straight making card for F as a turn card T for which there is a holdem hand H such that H didn't make a straight on F, but H makes a straight on F and T combined. A straight making rank is the rank of a straight making card. 
For example, 5 is a straight making rank for the flop 789 because the holdem hand 46 makes a straight on 7895, but not on 789. (The straight being made is 45678). All in all, 789 has four straight making ranks: 5,6, T and J. (4 and Q don't count because 56 (and TJ) were already making straights.)
EDIT: In the context of this question, the holdem hand must use both of its two cards to make a straight. So, for example, the hand K6 making a straight on a 7895 board doesn't count.
Now I have this conjecture:
(*) No flop has more than 6 straight making ranks.
Having examined a number of different flop types, I'm pretty sure that (*) is true. I'm just wondering if someone can come up with an elegant, succinct proof? (I admit that "elegant" and "succinct" are subjective terms).
 A: Not a solution, but some potentially helpful observations.
Since you're looking for a maximum numbers of SMRs, you should assume your flop is in the "middle" of the ranks.  Anything at an "edge" (near A or K) will only have fewer.
Let the flop ranks be $f_1, f_2, f_3$ in non-decreasing order, with differences $d_1 = f_2 - f_1$ and $d_2 = f_3-f_2$.  In order to make a straight, you need $\min(d_1,d_2) \leq 4$.  (The flop 2, 7, Q has no straight making ranks; the flop 2, 6, 10 has 6 SMRs.)
Looking at isolated cases (where one card is too far away to matter), a $d_i = 4$ has 3 SMRs (the ranks in between), $d_i = 3$ has 4 SMRs (the ranks in between, one below, and one above), $d_i = 2$ has 5 SMRs (the rank in between, two above and two below), $d_i = 1$ has 6 SMRs (ranks 3 below and 3 above), $d_i = 0$ has no SMRs.
To work through the combined cases, you can organize by $r = f_3 - f_1$.  If $r \geq 9$, you just have one of the isolated cases above.  For $r \leq 8$, you have to think through how the previous cases combine.  At least, using symmetry, you can assume $d_1 \geq d_2$.  There are 10 cases to work through, listed below as $r = d_1 + d_2$, two-part partitions of $r$ with largest part 4.  Note that any $d_i = 0$ reduces to one of the isolated cases above.
8 = 4+4 
7 = 4+3 
6 = 4+2 = 3+3
5 = 4+1 = 3+2
4 = 3+1 = 2+2
3 = 2+1
2 = 1+1
So not elegant, but it's not really that many cases to work through.
