Computing the first few terms of this sequence gives $a_1 = 2$, $a_2 = \frac{5}{2}$, $a_3 = \frac{10}{3}$, and $a_4 = \frac{17}{4}$, so it appears that $a_n = n + \frac{1}{n}$. Indeed, this is correct, because the recurrence relation is satisfied:
$$\frac{a_{n-1} + 1}{(n-1) + 1} = \frac{n - 1 + \frac{1}{n-1} + 1}{n} = 1 + \frac{1}{n(n-1)} = \frac{n-1}{n-1} + \frac{1}{n-1} = \frac{a_n - 1}{n-1}$$
Then
$$\begin{aligned} \sum_{n=1}^{100} a_n^2 &= \sum_{n=1}^{100} \left( n^2 + 2 + \frac{1}{n^2} \right) \\ &= \frac{100 \cdot 101 \cdot 201}{6} + 200 + r \\ &= \frac{2030100 + 1200}{6} + r \\ &= 338,550 + r, \end{aligned}$$
where $r = \sum_{n=1}^{100} \frac{1}{n^2}$. But by a telescoping sum argument,
$$1 < r < 1 + \sum_{n=2}^{100} \frac{1}{n(n-1)} = 2 - \frac{1}{100} < 2.$$
Thus $\sum_{n=1}^{100} a_n^2$ is between 338,551 and 338,552, and the requested greatest integer is 338,551.