What is the concentration of the original hydrochloric acid solution?

First, establish the reaction (as follows)

HCl(aq)   +     NaOH(aq) -----> NaCl(aq)  +   H2O(l)

Second, determine moles of NaOH used in the reaction
mole NaOH used = 24.16 mL (0.106 mol/1000 mL) = 0.00256 mol

Third, through stoichiometry from the above reaction, determine moles of HCl reacted
mole HCl reacted = 0.00256 mol NaOH(1mol HCl/1mol NaOH) = 0.00256 mol

Fourth, calculate for concentration for HCl

The concentration of a solution in molarity is defined as the mole of solute i.e mole of HCl per 1 liter of solution
thus,

Volume of HCl used in titration = 25 mL
M= 0.00256 mol/ (25/1000 L) = 0.1024 M

so you want to find out the concentration of HCl right? first do a balanced equation to see the mole ratios. If this is a one to one ratio. then: moles of HCl = moles of NaOH

so step 2:  

find moles of NaOH since you have the concentration and ml of it ( also convert ml to liters always!):
from:
M= mol / liter  

to:     Liters (M) = moles
 
so:       ( .024L NaOH )(0.106 M) = .002544 moles NaHO

next: since :     .002544 moles NaOH = moles of HCl

Then:   .002544 moles HCl /.025Liters = M of HCl  or .10176 M HCl (if you wanna round off then:    .102 M HCl

I hope I'm right, I just had an exam on this crap last week!