Answer: 25

Solution:

In $n$ years Jane's age is $25 + n,$ and Dick's age is its digit reversal. If Jane's future age is $10a + b,$ Dick's is $10b + a,$ which is larger exactly when $b \gt a.$ Conversely, every two-digit value of $25 + n$ with tens digit less than units digit yields exactly one valid pair: $n = 10a + b - 25$ and $d = 10b + a - n$ $= 25 + 9(b - a) \gt 25,$ so Dick is indeed older than Jane now.

So we count two-digit numbers that are at least $26$ and have tens digit less than units digit: $4$ starting with $2$ (namely $26$ through $29$), then $6,$ $5,$ $4,$ $3,$ $2,$ $1$ starting with $3$ through $8.$ The total is $4 + 6 + 5 + 4 + 3 + 2 + 1 = 25.$