Let $E$ be the set of all bijective mappings from $\mathbb R$ to $\mathbb R$ satisfying
\[f(t) + f^{-1}(t) = 2t, \qquad \forall t \in \mathbb R,\]
where $f^{-1}$ is the mapping inverse to $f$. Find all elements of $E$ that are monotonic mappings.
1. **Given Condition and Initial Analysis:**
   We are given that \( f \) is a bijective mapping from \(\mathbb{R}\) to \(\mathbb{R}\) satisfying:
   \[
   f(t) + f^{-1}(t) = 2t, \quad \forall t \in \mathbb{R},
   \]
   where \( f^{-1} \) is the inverse of \( f \). We need to find all elements of \( E \) that are monotonic mappings.

2. **Monotonicity and Inverse Function:**
   Since \( f \) is bijective and monotonic, \( f \) must be either strictly increasing or strictly decreasing. Similarly, \( f^{-1} \) must also be strictly monotonic in the same direction as \( f \) because the inverse of a strictly increasing function is strictly increasing, and the inverse of a strictly decreasing function is strictly decreasing.

3. **Sum of Functions:**
   Given the condition \( f(t) + f^{-1}(t) = 2t \), we can analyze the behavior of \( f \) and \( f^{-1} \). If \( f \) is strictly increasing, then \( f^{-1} \) is also strictly increasing, and their sum \( f(t) + f^{-1}(t) \) must be strictly increasing.

4. **Assumption and Contradiction:**
   Suppose \( f(x) - x \) is not constant. Let \( u \neq v \) such that \( f(u) - u = a \) and \( f(v) - v = b \) with \( a > b \). Using the given condition, we can derive:
   \[
   f(x) + f^{-1}(x) = 2x.
   \]
   It follows that:
   \[
   f(u + na) = u + (n+1)a \quad \text{and} \quad f(v + nb) = v + (n+1)b \quad \forall n \in \mathbb{Z}.
   \]
   Let \( n = \left\lfloor \frac{v - u}{a - b} \right\rfloor \). Then:
   \[
   n + 1 > \frac{v - u}{a - b} \geq n.
   \]
   This implies:
   \[
   \frac{v - u}{a - b} \geq n \implies v - u \geq n(a - b) \implies v + nb \geq u + na \implies f(v + nb) \geq f(u + na).
   \]
   Since \( f \) is increasing:
   \[
   v + (n+1)b \geq u + (n+1)a \implies \frac{v - u}{a - b} \geq n + 1.
   \]
   This leads to a contradiction.

5. **Conclusion:**
   Therefore, \( f(x) - x \) must be constant. Let \( f(x) = x + c \) for some constant \( c \). Substituting \( f(x) = x + c \) into the given condition:
   \[
   f(t) + f^{-1}(t) = 2t \implies (t + c) + (t - c) = 2t,
   \]
   which holds true for any real \( c \).

The final answer is \(\boxed{f(x) = x + c}\) for any real constant \( c \).