Okay, let's see. The problem is asking for the greatest possible value of the greatest common divisor (gcd) of two expressions: n + 7 and 2n + 1, where n is a positive integer. So, we need to find the maximum gcd that can occur for some n. Hmm, how do I approach this?

First, I remember that the gcd of two numbers can be found using the Euclidean algorithm. Maybe I can apply that here. Let's recall the Euclidean algorithm: to find gcd(a, b), we repeatedly subtract the smaller number from the larger one until we reach zero, and the last non-zero remainder is the gcd. Alternatively, we can use the property that gcd(a, b) = gcd(b, a mod b). Maybe that's useful here.

So, let's let a = 2n + 1 and b = n + 7. Then, gcd(a, b) = gcd(n + 7, (2n + 1) - 2*(n + 7)). Let me compute that. Let's do the subtraction step by step.

First, (2n + 1) minus 2*(n + 7) would be 2n + 1 - 2n - 14. Simplify that: 2n - 2n cancels out, 1 - 14 is -13. So, the remainder is -13. But since gcd is always non-negative, we can take the absolute value, so 13. Therefore, gcd(2n + 1, n + 7) = gcd(n + 7, 13). 

Wait, that's interesting. So, the gcd simplifies to gcd(n + 7, 13). Which means that the gcd is either 1 or 13, depending on whether n + 7 is a multiple of 13 or not. Since 13 is a prime number, its only divisors are 1 and itself. Therefore, the maximum possible value of the gcd is 13, provided that 13 divides n + 7. 

So, to have gcd(n + 7, 2n + 1) = 13, we need n + 7 ≡ 0 mod 13. That is, n ≡ -7 mod 13. Since n is a positive integer, we can choose n = 13k - 7 for some integer k ≥ 1. Let's check if this works. Let's take k = 1, then n = 13*1 - 7 = 6. Let's compute gcd(6 + 7, 2*6 + 1) = gcd(13, 13) = 13. Perfect, that works.

So, the maximum possible gcd is 13. Therefore, the answer should be 13. But let me verify this with another example to be sure. Let's take k = 2, then n = 13*2 - 7 = 26 - 7 = 19. Then gcd(19 + 7, 2*19 + 1) = gcd(26, 39). The gcd of 26 and 39. Let's compute that. 39 divided by 26 is 1 with remainder 13. Then gcd(26,13) is 13. So yes, that also gives 13. Another example: n = 32 (k=3), then n +7 =39, 2n +1=65. gcd(39,65)=13. Correct again.

What if I take a different n not congruent to -7 mod 13? Let's say n = 1. Then gcd(1 +7, 2*1 +1) = gcd(8,3) =1. If n=2, gcd(9,5)=1. If n=3, gcd(10,7)=1. If n=4, gcd(11,9)=1. n=5: gcd(12,11)=1. n=6: gcd(13,13)=13. So, as expected, only when n ≡6 mod13 (since 6 is equivalent to -7 mod13 because 6 +7=13) do we get the gcd of 13. Otherwise, the gcd is 1.

Therefore, the maximum possible gcd is indeed 13. There's no way to get a higher gcd because 13 is a prime number, and the next multiple would be 26, but 26 would require n +7 =26k, but then 2n +1 would be 2*(26k -7) +1=52k -14 +1=52k -13. Then gcd(26k,52k -13). Let's compute that. gcd(26k,52k -13). Using Euclidean algorithm: gcd(26k,52k -13) = gcd(26k,52k -13 -2*26k) = gcd(26k, -13). Absolute value is gcd(26k,13). Since 26 is a multiple of 13, gcd(26k,13) is 13 if k is not a multiple of 13, but if k is a multiple of 13, say k=13m, then n=26*13m -7=338m -7. Then n +7=338m, which is 26*13m. Then 2n +1=2*(338m -7)+1=676m -14 +1=676m -13. Then gcd(338m,676m -13). Let's compute that. gcd(338m,676m -13) = gcd(338m,676m -13 -2*338m) = gcd(338m, -13). Which is 13. So even if k is a multiple of 13, the gcd is still 13. Therefore, regardless of how large k is, the gcd remains 13. So 13 is the maximum possible value.

Therefore, the answer is 13. I think that's solid. Let me check once more. Suppose someone claims the gcd could be higher, say 26. Then 26 would have to divide both n +7 and 2n +1. If 26 divides n +7, then n ≡ -7 mod26. Then 2n +1 ≡ 2*(-7) +1 = -14 +1 = -13 mod26. So 2n +1 ≡ -13 mod26. But -13 mod26 is 13. So 2n +1 would be congruent to 13 mod26. Therefore, 26 divides 2n +1 -13, which is 2n -12. So 26 divides 2n -12. Then 13 divides n -6. But n ≡-7 mod26, which is n ≡19 mod26. Then n -6 ≡13 mod26. Therefore, 13 divides 13, which is true. So 26 divides 2n +1 if and only if 26 divides n +7 and 13 divides n -6. But n ≡-7 mod26 and n≡6 mod13. Wait, 6 mod13 is equivalent to 6 +13=19 mod26. So n ≡19 mod26. Let me check: n=19. Then n +7=26, 2n +1=39. gcd(26,39)=13, not 26. So even though 26 divides n +7, 2n +1 is 39, which is 13*3, so gcd is 13. Therefore, 26 can't be achieved. Hence, 13 is indeed the maximum possible.

Therefore, after all these checks and considerations, I'm confident that the greatest possible value of gcd(n +7, 2n +1) is 13.
Therefore, the greatest possible value of $\gcd(n + 7, 2n + 1)$ is $\boxed{13}$.
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