What is the greatest integer $n$ such that $$n\leq1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{2014}}?$$

$\text{(A) }31\qquad\text{(B) }59\qquad\text{(C) }74\qquad\text{(D) }88\qquad\text{(E) }112$
To find the greatest integer \( n \) such that 
\[ n \leq 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{2014}}, \]
we will approximate the sum using integrals and properties of the harmonic series.

1. **Approximate the sum using integrals:**
   Consider the function \( f(x) = \frac{1}{\sqrt{x}} \). The sum \( \sum_{k=1}^{2014} \frac{1}{\sqrt{k}} \) can be approximated by the integral:
   \[
   \int_{1}^{2014} \frac{1}{\sqrt{x}} \, dx.
   \]

2. **Evaluate the integral:**
   \[
   \int_{1}^{2014} \frac{1}{\sqrt{x}} \, dx = \left[ 2\sqrt{x} \right]_{1}^{2014} = 2\sqrt{2014} - 2\sqrt{1} = 2\sqrt{2014} - 2.
   \]

3. **Estimate \( \sqrt{2014} \):**
   \[
   \sqrt{2014} \approx \sqrt{2016} = \sqrt{4 \times 504} = 2\sqrt{504}.
   \]
   Since \( \sqrt{504} \approx 22.4 \) (because \( 22^2 = 484 \) and \( 23^2 = 529 \)), we have:
   \[
   \sqrt{2014} \approx 2 \times 22.4 = 44.8.
   \]

4. **Calculate the integral approximation:**
   \[
   2\sqrt{2014} - 2 \approx 2 \times 44.8 - 2 = 89.6 - 2 = 87.6.
   \]

5. **Add the initial term \( 1 \):**
   \[
   1 + \sum_{k=2}^{2014} \frac{1}{\sqrt{k}} \approx 1 + 87.6 = 88.6.
   \]

6. **Determine the greatest integer \( n \):**
   The greatest integer \( n \) such that \( n \leq 88.6 \) is \( 88 \).

Thus, the greatest integer \( n \) is \( \boxed{88} \).