Does C# allow double semicolon ; ; if so, are there any special ways?

No, a double semi-colon is never required. I'm slightly surprised that the compiler doesn't complain that it's an unreachable statement, but apparently it's legal. It won't do any harm, but equally it's not a good idea.

The empty statement is valid in all C-derived languages. The most common idiomatic use is in a for statement, e.g.:

for (; ; )
{
}

if their is a special use of it then show me where two semicolon is required to write a statement in c#

In the above example, two semicolons are required of course.

That a double ;; is allowed, is for historical reasons. It's is a hangover from C style languages (which C# is based on).

C & C++ have the concept of pre-processor macros which are replaced in the code before the code is compiled e.g. trivial example AddAndSquare is a macro, not a function

#define AddAndSquare(X,Y) (X+Y)*(X+Y)
int Foo() {
   int a = 1, b = 2;
   return AddAndSquare(a, b);
}

goes to the compiler as

int Foo() {
   int a = 1, b = 2;
   return (A+B)*(A+B);
}

You can redefine macros to be different to their initial definition, also you can redefine them so they don't exist at all.

Given an assertion macro #define ASSERT(c) if(!c) throw new AssertionFailedException() you can have your coded littered with ASSERT statements.

void Foo(int x) {
    int y = x + 2;
    ASSERT(y != 0);
   int z = x / y;
    . . . .
}

Now consider that you only want the asserts in debug builds, but not in release builds, for release you redefine the macro to be empty (literally #define ASSERT). Now when Foo goes to the compiler for a release build, it looks like this

void Foo(int x) {
    int y = x + 2;
    ;
   int z = x / y;
    . . . .
}

There's now an empty statement where the ASSERT was, because there may or may not be a statement there (depending on build configuration), the compiler needs to be able to handle an empty statement.

Why this convention was kept in C# where there are nothing like C macros, I have no idea, but possibly because it causes little or no harm.

I would guess that multiple ; are elided by the compiler before it starts parsing code, therefore your unreachable ; is ignored by the compiler.