Several nice solutions - a sample of which are listed below as they differ slightly. You might like to decide which one makes most sense to you.

Submitters: Michael of Spalding Grammer, Mary of Birchwood Community High School, Andrei of School 205, Bucharest, Tan Chor or the Raffles Institution, Sana, Jenny, Chris and Rosion of Madras College, St. Andrews and Chen of The Chinese High School, Singapore .

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Let 2.5252... be x

      x = 2.5252...
100x = 252.5252...
  99x = 250
So  x = 250 / 99

Hence, sum of numerator and denominator = 250 + 99 = 349

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The sum of the denominator and numerator is 349.

We divided 5 by 1 which equals 5.

We divided 5 by 2 which equals 2.5 or 2 1/2.

So we knew that 5 had to be divided by a number between 1 and 2 to get 2.5252525…

Using trial and error we finally divided 5 by 1.98 or 1 49/50.

5 / 1.98 = 2.52525252525252525252525…

or

2 52/99

We changed 2 52/99 into a top heavy fraction:

2 52/99 = (2 x 99) + 52 / 99

= 198 + 52 / 99

= 250 / 99

Then we added the denominator to the numerator:

99 + 250 = 349

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Given a recurring decimal 0.xyxyxyxy... where x and y represent digits, it can be shown that 0.xyxyxyxy... = xy/99

        100 * 0.xyxyxyxyxy.... = xy.xyxyxyxy...
Thus, 99 * 0.xyxyxyxy... .... = xy.xyxyxyxy... - 0.xyxyxyxy... = xy

Dividing 99 from both sides, we see that 0.xyxyxyxy... is equal to xy/99.

Having derived this formula, 2.525252... = 2 52/99 = 250/99. Adding the denominator and numberator, the final answer is 349.

Extending the above formula for all recurring decimals, it can again be derived that 0.a1a2a3...ax where a1 to ax represent digits (1 to x are in subscript) and a1a2a3...ax recurs, the recurring decimal is equal to a1a2a3a4...a x/[(10x) - 1]