Solution:

Let the initial number of eggs = $x$.

First sale (to Anurag):
Sells half of the eggs plus half an egg:
$$\text{Eggs sold}=\frac{x}{2}+\frac{1}{2}$$ Remaining eggs: $$x-(\frac{x}{2}+\frac{1}{2})=\frac{x}{2}-\frac{1}{2}$$
Second sale (to Deepak):
Sells half of remaining eggs plus half an egg:
Remaining eggs: $$(\frac{x}{2}-\frac{1}{2})-(\frac{1}{2}(\frac{x}{2}-\frac{1}{2})+\frac{1}{2})=\frac{x}{4}-\frac{5}{4}$$
Third sale (to Shivani):
Sells half of remaining eggs plus half an egg:
Remaining eggs: $$(\frac{x}{4}-\frac{5}{4})-(\frac{1}{2}(\frac{x}{4}-\frac{5}{4})+\frac{1}{2})=\frac{x-9}{8}$$
Given that after all sales, 7 eggs are left: $$\frac{x-9}{8}=7$$ Multiplying both sides by 8: $$x-9=56$$ $$x=65$$
Verification:
Since dealer never broke an egg, and he must sell whole number of eggs each time:

Checking for $x=65$:
- After first sale: Remaining eggs = 32
- After second sale: Remaining eggs = 16
- After third sale: Remaining eggs = 7 (Not matching without breaking eggs). ❌

Checking for $x=63$:
- After first sale: Sells $\frac{63}{2}+\frac{1}{2}=32$ eggs, Remaining = 31 eggs ✅
- After second sale: Sells $\frac{31}{2}+\frac{1}{2}=16$ eggs, Remaining = 15 eggs ✅
- After third sale: Sells $\frac{15}{2}+\frac{1}{2}=8$ eggs, Remaining = 7 eggs ✅

Thus, the dealer originally had: $$\boxed{{\displaystyle 63\text{eggs}}}$$