Randomly Choose from list but meet conditions

To create some data for testing we get even numbers <100 and insert one odd number:

d = Table[i, {i, 0, 99, 2}];
d = Insert[d, 33, 18];

Now we can pick out a sequence of 3 integers in a row by:

SequenceCases[d, {x_, y_, z_} /; z == (y + 1) == x + 2, 1]
(*  {{32, 33, 34}}  *)

If you only want the first number, you would say:

SequenceCases[d, {x_, y_, z_} /; z == (y + 1) == x + 2 -> x, 1][[1]]
(* 32 *)

Here's a way that constructs a set of candidates by sorting the list and then breaking it into runs of consecutive elements and only keeping elements with enough successors to be viable. This is probably a good approach if you are going to be making multiple draws.

RandomSeed[1337];
test = RandomInteger[{0, 99}, 95]
(* {37, 84, 80, 98, 26, 32, 51, 65, 19, 33, 10, 88, 25, 13,
    77, 68, 20, 39, 27, 83, 89, 75, 78, 87, 22, 58, 94, 49, 
    70, 73, 60, 44, 59, 86, 30, 12, 90, 5, 55, 63, 38, 35, 
    72, 81, 31, 36, 29, 93, 95, 8, 34, 15, 9, 42, 52, 50, 
    48, 4, 92, 71, 56, 2, 69, 54, 74, 91, 45, 64, 0, 82, 
    96, 85, 16, 6, 7} *) 
runs = Split[Sort[test], Subtract /* EqualTo[-1]];

Only runs of length $ > 3 $ will contain elements $ a $ such that $ a + 1 $, $ a + 2 $, and $ a + 3 $ are present:

longRuns = Select[runs, Length /* GreaterThan[3]];

Drop elements that don't have enough successors and flatten.

candidates = Flatten[Drop[#, -3] & /@ longRuns]
(* {4, 5, 6, 7, 29, 30, 31, 32, 33, 34, 35, 36, 48, 49, 68, 
    69, 70, 71, 72, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89,
    90, 91, 92, 93} *)

Now you can make random draws.

RandomChoice[candidates, 10]
(* {34, 85, 89, 90, 4, 81, 83, 29, 93, 85} *)

First generate a random sample of integers between 0 and 99 (and this can be with or without replacement - but I've used "without replacement" in the example):

nOriginal = 60;
original = RandomSample[Range[0, 99], nOriginal]

Now find all of the available starting numbers (a) that have a+1, a+2, and a+3 available:

available = Select[original,
  MemberQ[original, # + 1] && MemberQ[original, # + 2] && MemberQ[original, # + 3] &]

Now you can select from available with or without replacement as appropriate for your sampling objective.

This approach should work whether or not the original list has duplicate numbers or not. (However, this approach is much slower than that of @Pillsy.)

Since the input list not too large, we can also play with alternative approaches, among them RelationGraph:

SeedRandom[1]
length = 60;
inputlist = RandomSample[Range[0, 99], length]

{80, 14, 0, 67, 3, 65, 23, 68, 74, 15, 24, 4, 83, 70, 1, 30, 48, 25, 44, 73,
 69, 56, 47, 28, 92, 26, 75, 10, 43, 33, 81, 18, 38, 29, 84, 17, 27, 85, 5, 40,
 82, 22, 2, 39, 36, 20, 99, 46, 52, 54, 35, 9, 63, 42, 95, 89, 98, 49, 64, 45}

rg = RelationGraph[#2 == # + 1 &, inputlist, ImageSize -> Large, 
  VertexSize -> .7, VertexStyle -> White, 
  VertexLabels -> Placed["Name", Center]]

candidates = VertexList[rg, _?(Length[VertexOutComponent[rg, #]] >= 4 &)]

 {80, 0, 67, 23, 24, 1, 25, 44, 26, 43, 81, 27, 82, 22, 2, 46, 42, 45}

HighlightGraph[rg, Style[#, Orange] & /@ candidates]

Use RandomSample or RandomChoice to pick any number of elements from candidates:

SeedRandom[123]
randomselection = RandomSample[candidates, 5]

{2, 80, 67, 25, 44}

HighlightGraph[rg, 
 Join[Style[# + Range[3], Yellow] & /@ randomselection, List /@ randomselection]]