Decompose 96 into two positive integer factors whose sum of squares is 208.
As the product of the two factors is 96, they are divisors of 96. Decomposing 96 into prime factors, we find $96=2^{5} \times 3$, therefore, its divisors are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, and 96.

The divisors 96, 48, 32, 24, and 16 do not work, as their squares are already greater than 208, leaving 1, 2, 3, 4, 6, 8, and 12, whose squares are 1, 4, 9, 16, 36, 64, and 144.

It is now easy to see that the only possibility is $64+144=208$. Since $8 \times 12=96$, the numbers are 8 and 12.