Suppose they're ordered $a_1\leq a_2\leq \ldots\leq a_{2004}$. Since $\sum_{i=1}^{2003}i^2=2003\cdot 334\cdot4007>2003^2$, we have that $\sum_{i=1}^{2004}\left(\frac{5(i-1)}{2003}\right)^2>25$. Thus there's some $a_i$ with $a_i<\left(\frac{5(i-1)}{2003}\right)^2$; let $i$ be the least such index. Clearly $i>1$; then $\frac{5(i-2)}{2003}\leq\sqrt{a_{i-1}}\leq\sqrt{a_i}<\frac{5(i-1)}{2003}$. Taking the differences of the two outer inequalities gives the required result. There was a fair amount of boiling down there. Why would they choose $\frac{5}{2003}$? They could've chosen any number at least $\frac{5}{\sqrt{2003\cdot334\cdot4007}}\approx 0.00009657097$, whereas $\frac{5}{2003}\approx0.00249626$. They were off by about an order of magnitude.

I think that there was a typo in the questionnaire... I think that they meant that all of $a_1$, ..., $a_{2004}$ are less than or equal to 25, not the sum of them...

Here is a faster solution: Assume the hypothesis is false, and $|\sqrt{a_i}-\sqrt{a_j}|>5/2003$ for all $j\neq i$. Now, let $a_1<a_2<\cdots<a_{2004}$ without loss of generality. We have $\sqrt{a_{k+1}}-\sqrt{a_k}>5/2003$ for all $k$. Summing these inequalities up for $k=1,2,\dots,2003$, we get that $\sqrt{a_{2004}}-\sqrt{a_1}>5$. This yields $a_{2004}>25$, which contradicts with $a_1+\cdots+a_{2004}\leqslant 25$.