We can solve this problem by using a simple reduction.

Let's assume that you have a decider named '2014' which takes input as the description of a Turing machine i.e <M>, and accepts <M> if M accepts any string of length 2014 otherwise it rejects <M>.

clearly, The language accepted by '2014' is all the <M>'s which accepts at least one string of length 2014. (Problem statement).

Now say if you really have a decider like '2014', I'll use it to solve the halting problem.

So, I want to build a decider D for language A = {<M, w> | M is a Turing machine which accepts string w} i.e a decider for the halting problem.

My decider D takes input <M,w> and creates another TM D'.

D' does the following things:

Run M on w.

If M halts on w then D' accepts its input.

Clearly, D' accepts all the inputs if M halts on w. So, D' accepts a string of length 2014 if and only if M halts on w.

D' accepts no string if M does not halt on w.

Give <D'> as an input to '2014' .

if '2014' accepts <D'> then D accepts <M,w>

else D rejects <M,w>

Clearly, we can solve the halting problem if we have decider like '2014', but, we know that HP is Undecidable. So '2014' Must be undecidable.

Please check if this is a valid reduction or not. @Arjun sir.