Numbers $1$ through $2014$ are written on a board. A valid operation is to erase two numbers $a$ and $b$ on the board and replace them with the greatest common divisor and the least common multiple of $a$ and $b$.

Prove that, no matter how many operations are made, the sum of all the numbers that remain on the board is always larger than $2014$ $\times$ $\sqrt[2014]{2014!}$
1. **Claim 1: \(\left( \dfrac{2015}{2} \right)^{2014} > 2014!\)**

   By the Arithmetic Mean-Geometric Mean Inequality (AM-GM), we know that for any set of positive numbers \(a_1, a_2, \ldots, a_n\),
   \[
   \frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}
   \]
   with equality if and only if \(a_1 = a_2 = \cdots = a_n\).

   Applying AM-GM to the numbers \(1, 2, \ldots, 2014\), we get:
   \[
   \frac{1 + 2 + \cdots + 2014}{2014} \geq \sqrt[2014]{1 \cdot 2 \cdot \cdots \cdot 2014}
   \]
   The sum of the first 2014 positive integers is:
   \[
   1 + 2 + \cdots + 2014 = \frac{2014 \cdot 2015}{2}
   \]
   Therefore,
   \[
   \frac{\frac{2014 \cdot 2015}{2}}{2014} = \frac{2015}{2} \geq \sqrt[2014]{2014!}
   \]
   Raising both sides to the power of 2014, we obtain:
   \[
   \left( \frac{2015}{2} \right)^{2014} \geq 2014!
   \]

2. **Claim 2: \(\gcd(a,b) + \text{lcm}(a,b) \geq a + b\)**

   Without loss of generality, assume \(a \leq b\). We know that:
   \[
   \gcd(a, b) \leq a
   \]
   and
   \[
   \text{lcm}(a, b) = \frac{ab}{\gcd(a, b)}
   \]
   Since \(b = \frac{ab}{a}\), we can define the function:
   \[
   f(x) = x + \frac{ab}{x}
   \]
   We need to show that:
   \[
   f(\gcd(a, b)) \geq f(a)
   \]
   The function \(f(x)\) is decreasing for \(x \in (0, \sqrt{ab}]\). Since \(\gcd(a, b) \leq a\) and both \(\gcd(a, b)\) and \(a\) are in the interval \((0, \sqrt{ab}]\), we have:
   \[
   f(\gcd(a, b)) \geq f(a)
   \]
   Therefore,
   \[
   \gcd(a, b) + \text{lcm}(a, b) \geq a + b
   \]

3. **Proof:**

   Observe that the sum of the numbers on the board does not decrease by our second claim. Initially, the sum of the numbers is:
   \[
   1 + 2 + \cdots + 2014 = \frac{2014 \cdot 2015}{2}
   \]
   By the first claim, we have:
   \[
   \left( \frac{2015}{2} \right)^{2014} > 2014!
   \]
   Therefore,
   \[
   \frac{2015}{2} > \sqrt[2014]{2014!}
   \]
   Multiplying both sides by 2014, we get:
   \[
   \frac{2014 \cdot 2015}{2} > 2014 \cdot \sqrt[2014]{2014!}
   \]
   Thus, the sum of all the numbers that remain on the board is always larger than:
   \[
   2014 \times \sqrt[2014]{2014!}
   \]

\(\blacksquare\)

The final answer is \( \boxed{ 2014 \times \sqrt[2014]{2014!} } \)