Let $n$ be an integer greater than $1$.Find all pairs of integers $(s,t)$ such that equations:
$x^n+sx=2007$
and
$x^n+tx=2008$
have at least one common real root.
1. Given the equations \( x^n + sx = 2007 \) and \( x^n + tx = 2008 \), we need to find pairs \((s, t)\) such that these equations have at least one common real root.

2. Let \( x \) be the common real root. Then:
   \[
   x^n + sx = 2007 \quad \text{and} \quad x^n + tx = 2008
   \]

3. Subtract the first equation from the second:
   \[
   (x^n + tx) - (x^n + sx) = 2008 - 2007
   \]
   Simplifying, we get:
   \[
   (t - s)x = 1
   \]
   Therefore:
   \[
   x = \frac{1}{t - s}
   \]

4. Substitute \( x = \frac{1}{t - s} \) into the second equation \( x^n + tx = 2008 \):
   \[
   \left( \frac{1}{t - s} \right)^n + t \left( \frac{1}{t - s} \right) = 2008
   \]
   Multiplying through by \( (t - s)^n \):
   \[
   1 + t(t - s)^{n-1} = 2008(t - s)^n
   \]

5. Since \( t - s \) must be an integer, and \( t - s \) divides 1, the possible values for \( t - s \) are \( 1 \) and \( -1 \).

6. Case 1: \( t - s = 1 \)
   \[
   x = \frac{1}{1} = 1
   \]
   Substitute \( x = 1 \) into the equations:
   \[
   1^n + t \cdot 1 = 2008 \implies t + 1 = 2008 \implies t = 2007
   \]
   \[
   1^n + s \cdot 1 = 2007 \implies s + 1 = 2007 \implies s = 2006
   \]
   Therefore, one pair is \( (s, t) = (2006, 2007) \).

7. Case 2: \( t - s = -1 \)
   \[
   x = \frac{1}{-1} = -1
   \]
   Substitute \( x = -1 \) into the equations:
   \[
   (-1)^n + t \cdot (-1) = 2008 \implies (-1)^n - t = 2008 \implies t = -2008 - (-1)^n
   \]
   \[
   (-1)^n + s \cdot (-1) = 2007 \implies (-1)^n - s = 2007 \implies s = -2007 - (-1)^n
   \]

   - If \( n \) is even (\( n = 2k \)):
     \[
     (-1)^{2k} = 1
     \]
     \[
     t = -2008 - 1 = -2009
     \]
     \[
     s = -2007 - 1 = -2008
     \]
     Therefore, another pair is \( (s, t) = (-2008, -2009) \).

   - If \( n \) is odd (\( n = 2k + 1 \)):
     \[
     (-1)^{2k+1} = -1
     \]
     \[
     t = -2008 - (-1) = -2007
     \]
     \[
     s = -2007 - (-1) = -2006
     \]
     Therefore, another pair is \( (s, t) = (-2006, -2007) \).

The final answer is \( \boxed{ (2006, 2007) } \), \((-2008, -2009)\), and \((-2006, -2007)\).