Let $a < b < c < d < e$ be real numbers. We calculate all possible sums in pairs of these 5 numbers. Of these 10 sums, the three smaller ones are 32, 36, 37, while the two larger ones are 48 and 51. Determine all possible values ​​that $e$ can take.
1. We start by noting that we have five real numbers \(a < b < c < d < e\) and we are given the sums of pairs of these numbers. Specifically, the three smallest sums are 32, 36, and 37, and the two largest sums are 48 and 51.

2. Let's denote the sums of pairs as follows:
   \[
   a + b = 32, \quad a + c = 36, \quad \text{and either } a + d \text{ or } b + c = 37.
   \]

3. First, consider the case where \(a + d = 37\):
   \[
   a + b = 32, \quad a + c = 36, \quad a + d = 37.
   \]
   Subtracting the first equation from the second:
   \[
   (a + c) - (a + b) = 36 - 32 \implies c - b = 4.
   \]
   Subtracting the first equation from the third:
   \[
   (a + d) - (a + b) = 37 - 32 \implies d - b = 5.
   \]
   Subtracting the second equation from the third:
   \[
   (a + d) - (a + c) = 37 - 36 \implies d - c = 1.
   \]
   Therefore, we have:
   \[
   c = b + 4, \quad d = b + 5, \quad d = c + 1.
   \]
   This is consistent, so we proceed with these values.

4. Next, consider the largest sums:
   \[
   d + e = 51, \quad c + e = 48.
   \]
   Subtracting the second equation from the first:
   \[
   (d + e) - (c + e) = 51 - 48 \implies d - c = 3.
   \]
   However, we previously found \(d - c = 1\), which is a contradiction. Therefore, \(a + d = 37\) is not possible.

5. Now consider the case where \(b + c = 37\):
   \[
   a + b = 32, \quad a + c = 36, \quad b + c = 37.
   \]
   Subtracting the first equation from the third:
   \[
   (b + c) - (a + b) = 37 - 32 \implies c - a = 5.
   \]
   Subtracting the second equation from the third:
   \[
   (b + c) - (a + c) = 37 - 36 \implies b - a = 1.
   \]
   Therefore, we have:
   \[
   b = a + 1, \quad c = a + 5.
   \]

6. Next, consider the largest sums:
   \[
   d + e = 51, \quad c + e = 48.
   \]
   Subtracting the second equation from the first:
   \[
   (d + e) - (c + e) = 51 - 48 \implies d - c = 3.
   \]
   Therefore, we have:
   \[
   d = c + 3 = (a + 5) + 3 = a + 8.
   \]

7. Now, we solve for \(e\):
   \[
   c + e = 48 \implies (a + 5) + e = 48 \implies e = 43 - a.
   \]
   \[
   d + e = 51 \implies (a + 8) + e = 51 \implies e = 43 - a.
   \]
   Both equations for \(e\) are consistent.

8. We now have:
   \[
   a + b = 32 \implies a + (a + 1) = 32 \implies 2a + 1 = 32 \implies 2a = 31 \implies a = 15.5.
   \]
   \[
   b = a + 1 = 16.5, \quad c = a + 5 = 20.5, \quad d = a + 8 = 23.5, \quad e = 43 - a = 27.5.
   \]

Thus, the only possible value for \(e\) is \(27.5\).

The final answer is \(\boxed{27.5}\)