Assume the contrary then there are $\frac{49\times 48}{2}=1176$ ways of choosing a pairs of numbers.Also the sums of any pairs can range from $3$ to $599+600=1199$ so there are $1197$ possible sums and $1176$ pairs so by PHP two pairs must have the same sum.A contradiction.

leibnitz wrote: Assume the contrary then there are $\frac{49\times 48}{2}=1176$ ways of choosing a pairs of numbers.Also the sums of any pairs can range from $3$ to $599+600=1199$ so there are $1197$ possible sums and $1176$ pairs so by PHP two pairs must have the same sum.A contradiction. That is false, since number of pairs is less than possible sums.