We add initial term $S_{0}=0$ to the sequence $S_{1}, S_{2}, \ldots, S_{100}$ and consider all the terms $S_{n_{0}}<S_{n_{1}}<\ldots$ that are perfect squares: $S_{n_{k}}=m_{k}^{2}$ (in particular, $n_{0}=m_{0}=0$). Since $S_{100}=5050<72^{2}$, all the numbers $m_{k}$ do not exceed 71. If $m_{k+1}=m_{k}+1$ the difference $S_{n_{k+1}}-S_{n_{k}}=2 m_{k}+1$ is odd, and an odd number must occur among the numbers $a_{n_{k}+1}, \ldots, a_{n_{k+1}}$. There are only 50 odd numbers less than 100, so at most 50 differences $m_{k+1}-m_{k}$ equal 1. If there is 61 perfect squares in the original sequence, then $m_{61}=(m_{61}-m_{60})+(m_{60}-m_{59})+\ldots+(m_{1}-m_{0}) \geqslant 50+11 \cdot 2=72$, a contradiction. It remains to give an example of sequence containing 60 perfect squares. Let $a_{i}=2 i-1$ for $1 \leqslant i \leqslant 50$, then we use all the odd numbers and $S_{i}=i^{2}$. Further, let $a_{51+4 i}=2+8 i, a_{52+4 i}=100-4 i, a_{53+4 i}=4+8 i, a_{54+4 i}=98-4 i$ for $0 \leqslant i \leqslant 7$; thus we use all the even numbers between 70 and 100 and all the numbers between 2 and 60 that leave the remainder 2 or 4 when divided by 8. For $0 \leqslant i \leqslant 7$ we have $S_{54+4 i}-S_{50+4 i}=204+8 i$, and $S_{54+4 i}=(52+2 i)^{2}$. Finally, let the last 18 terms of the sequence be $30,40,64,66,68,6,8,14,16,32,38,46,54,62,22,24,48,56$. This gives $S_{87}=66^{2}+2 \cdot 134=68^{2}, S_{96}=70^{2}$.