We determine the values of $\lambda$ for which Alice has a winning strategy and those for which Bazza has a winning strategy. Let $S_n = \sum_{i=1}^n x_i$ and $Q_n = \sum_{i=1}^n x_i^2$. Alice (A) plays at odd $n$, ensuring $x_n \geq 0$ and $S_n \leq \lambda n$. Bazza (B) plays at even $n$, ensuring $x_n \geq 0$ and $Q_n \leq n$. The critical value for $\lambda$ is $\frac{1}{\sqrt{2}}$.

\textbf{Case 1: $0 < \lambda < \frac{1}{\sqrt{2}}$. Bazza has a winning strategy.}

Let $\delta = \sqrt{2} - 2\lambda$. Since $\lambda < \frac{1}{\sqrt{2}}$, we have $\delta > 0$.

Bazza's strategy (B-MaxQ) is to ensure $Q_{2k} = 2k$ at every turn $n = 2k$. This requires choosing $x_{2k} = \sqrt{2k - Q_{2k-1}}$. This is feasible if $Q_{2k-1} \leq 2k$.

Let $C_k$ be the budget available to Alice at the start of turn $2k-1$: $C_k = \lambda(2k - 1) - S_{2k-2}$ (with $S_0 = 0$). Alice must choose $x_{2k-1} \in [0, C_k]$. If $C_k < 0$, Alice loses immediately.

We analyze the evolution of $C_k$, assuming the game continues and Bazza follows B-MaxQ. $C_{k+1} = \lambda(2k+1) - S_{2k} = C_k + 2\lambda - (x_{2k-1} + x_{2k})$.

If Bazza successfully follows B-MaxQ up to turn $2k$, then $Q_{2k} = 2k$ and $Q_{2k-2} = 2k - 2$. Thus, $x_{2k-1}^2 + x_{2k}^2 = Q_{2k} - Q_{2k-2} = 2$. Since $x_i \geq 0$, $(x_{2k-1} + x_{2k})^2 = x_{2k-1}^2 + x_{2k}^2 + 2x_{2k-1}x_{2k} \geq 2$, so $x_{2k-1} + x_{2k} \geq \sqrt{2}$.

Therefore, $C_{k+1} \leq C_k + 2\lambda - \sqrt{2} = C_k - \delta$.

We must verify that B-MaxQ is always feasible as long as the game continues (i.e., $C_k \geq 0$). We proceed by induction. $C_1 = \lambda$. Since $\delta > 0$, if $C_k \geq 0$, the sequence $C_k$ is strictly decreasing. Thus $C_k \leq C_1 = \lambda$. Since $\lambda < 1/\sqrt{2}$. Alice must choose $x_{2k-1} \leq C_k < 1/\sqrt{2}$. If Bazza maintained $Q_{2k-2} = 2k - 2$, then $Q_{2k-1} = Q_{2k-2} + x_{2k-1}^2 = 2k - 2 + x_{2k-1}^2 < 2k - 2 + 1/2 = 2k - 3/2$. Since $Q_{2k-1} < 2k$, Bazza can choose $x_{2k}$ to achieve $Q_{2k} = 2k$. B-MaxQ is always feasible.

Since $C_{k+1} \leq C_k - \delta$, the budget decreases by at least $\delta$ in each round pair. $C_k \leq C_1 - (k - 1)\delta = \lambda - (k - 1)\delta$. Since $\lambda$ is fixed and $\delta > 0$, there exists an integer $K$ such that $(K - 1)\delta > \lambda$. For this $K$, $C_K < 0$. At turn $2K - 1$, Alice needs to choose $x_{2K-1} \geq 0$ such that $x_{2K-1} \leq C_K$. Since $C_K < 0$, no such choice exists. Bazza wins.

\textbf{Case 2: $\lambda > \frac{1}{\sqrt{2}}$. Alice has a winning strategy.}

Consider the function $h(K) = \frac{K\sqrt{2}}{2K-1}$ for $K \geq 1$. $h(K)$ is strictly decreasing and $\lim_{K \to \infty} h(K) = 1/\sqrt{2}$. Since $\lambda > 1/\sqrt{2}$, there exists an integer $K \geq 1$ such that $\lambda > h(K)$. This implies $L = \lambda(2K - 1) > K\sqrt{2}$.

Alice's strategy (A-Spike-K): Play $x_{2i-1} = 0$ for $i = 1, \ldots, K - 1$. At turn $2K - 1$, play the maximum possible value.

First, we verify the feasibility. For $i < K$, Alice plays $x_{2i-1} = 0$. She needs $S_{2i-1} = S_{2i-2} \leq \lambda(2i - 1)$. Bazza is constrained by $Q_{2i-2} \leq 2(i - 1)$. By the QM-AM inequality (or Cauchy-Schwarz), $S_{2i-2} \leq \sqrt{(i - 1)Q_{2i-2}} \leq \sqrt{(i - 1)2(i - 1)} = (i - 1)\sqrt{2}$. We check the constraint: $(i - 1)\sqrt{2} \leq \lambda(2i - 1)$, or $\lambda \geq \frac{(i-1)\sqrt{2}}{2i-1}$. The RHS is an increasing sequence converging to $1/\sqrt{2}$. Since $\lambda > 1/\sqrt{2}$, the strategy is feasible.

Now we analyze the outcome. Let $N = K - 1$. Bazza has made $N$ moves $y_i = x_{2i}$ ($i = 1, \ldots, N$). At turn $2K - 1$, Alice plays $x_{2K-1} = L - S_{2N}$. Since $S_{2N} \leq N\sqrt{2}$ and $L > K\sqrt{2} = (N + 1)\sqrt{2}$, $x_{2K-1} > \sqrt{2} > 0$.

Alice wins if Bazza cannot move at turn $2K$, i.e., $Q_{2K-1} > 2K$. $Q_{2K-1} = Q_{2N} + (L - S_{2N})^2$.

Bazza aims to minimize this quantity subject to his constraints: $y_i \geq 0$ and $\sum_{j=1}^i y_j^2 \leq 2i$. These constraints imply $Q_{2N} \leq 2N$, and consequently $S_{2N} \leq N\sqrt{2}$.

Let $F(y) = Q_{2N}(y) + (L - S_{2N}(y))^2$. Consider the strategy $y^* = (\sqrt{2}, \ldots, \sqrt{2})$. This is feasible for Bazza as $\sum_{i=1}^i (\sqrt{2})^2 = 2i$. Let $S^* = N\sqrt{2}$ and $Q^* = 2N$.

Let $y$ be any feasible strategy for Bazza. Let $\Delta S = S^* - S_{2N}(y) \geq 0$. We compare $F(y)$ with $F(y^*)$. We use the identity $\sum (y_i - \sqrt{2})^2 = Q_{2N}(y) - 2\sqrt{2}S_{2N}(y) + 2N$. $Q_{2N}(y) - Q^* = Q_{2N}(y) - 2N = \sum (y_i - \sqrt{2})^2 + 2\sqrt{2}S_{2N}(y) - 4N$. $2\sqrt{2}S_{2N}(y) = 2\sqrt{2}(S^* - \Delta S) = 4N - 2\sqrt{2}\Delta S$. $Q_{2N}(y) - Q^* = \sum (y_i - \sqrt{2})^2 - 2\sqrt{2}\Delta S$.

$F(y) - F(y^*) = Q_{2N}(y) - Q^* + (L - S_{2N}(y))^2 - (L - S^*)^2$. $(L - S_{2N}(y))^2 = (L - (S^* - \Delta S))^2 = (L - S^*)^2 + 2(L - S^*)\Delta S + (\Delta S)^2$.

$F(y) - F(y^*) = (\sum (y_i - \sqrt{2})^2 - 2\sqrt{2}\Delta S) + 2(L - S^*)\Delta S + (\Delta S)^2$. $F(y) - F(y^*) = \sum (y_i - \sqrt{2})^2 + 2(L - S^* - \sqrt{2})\Delta S + (\Delta S)^2$.

By the choice of $K$, $L > K\sqrt{2} = (N + 1)\sqrt{2} = S^* + \sqrt{2}$. Let $\epsilon = L - S^* - \sqrt{2} > 0$. $F(y) - F(y^*) = \sum (y_i - \sqrt{2})^2 + 2\epsilon\Delta S + (\Delta S)^2$. Since all terms are non-negative, $F(y) \geq F(y^*)$. The minimum value of $Q_{2K-1}$ is $F(y^*)$.

$Q_{2K-1} \geq F(y^*) = 2N + (L - N\sqrt{2})^2 = 2N + (\sqrt{2} + \epsilon)^2$. Since $\epsilon > 0$, $(\sqrt{2} + \epsilon)^2 > 2$. $Q_{2K-1} > 2N + 2 = 2K$. Bazza cannot move at turn $2K$. Alice wins.

\textbf{Case 3: $\lambda = \frac{1}{\sqrt{2}}$. Neither player has a winning strategy.}

We show that both players have a strategy to ensure the game continues forever (a draw).

\begin{enumerate}
    \item Alice's drawing strategy (A-Zero): $x_{2k-1} = 0$ for all $k$. We verify the game continues forever. Alice's feasibility at turn $2k - 1$: We need $S_{2k-2} \leq \lambda(2k - 1)$. Bazza maximizes $S_{2k-2}$ subject to $Q_{2k-2} \leq 2k - 2$, achieving at most $(k-1)\sqrt{2}$. We check: $(k - 1)\sqrt{2} \leq \frac{1}{\sqrt{2}}(2k - 1) \iff 2k - 2 \leq 2k - 1$. True. Bazza's survival at turn $2k$: We need $Q_{2k-1} \leq 2k$. $Q_{2k-1} = Q_{2k-2} \leq 2k - 2 < 2k$. Bazza survives. Alice's survival at turn $2k + 1$: We need $S_{2k} \leq \lambda(2k + 1)$. Bazza maximizes $S_{2k}$ subject to $Q_{2k} \leq 2k$, achieving at most $k\sqrt{2}$. We check: $k\sqrt{2} \leq \frac{1}{\sqrt{2}}(2k + 1) \iff 2k \leq 2k + 1$. True. The game continues forever. Bazza cannot win.
    \item Bazza's drawing strategy (B-MaxQ): $Q_{2k} = 2k$. We verify the game continues forever. Bazza's feasibility (survival). As shown in Case 1, if Bazza follows B-MaxQ, $S_{2k-2} \geq (k-1)\sqrt{2}$. Alice's budget $C_k = \lambda(2k-1) - S_{2k-2}$. $C_k \leq \frac{1}{\sqrt{2}}(2k-1) - (k-1)\sqrt{2} = \frac{2k-1-2(k-1)}{\sqrt{2}} = \frac{1}{\sqrt{2}}$. Alice must choose $x_{2k-1} \leq 1/\sqrt{2}$. Then $Q_{2k-1} = 2k - 2 + x_{2k-1}^2 \leq 2k - 2 + 1/2 < 2k$. B-MaxQ is feasible. Bazza survives.

    Alice's survival. We must show $C_k > 0$ for all $k$. $C_1 = 1/\sqrt{2} > 0$. $C_{k+1} = C_k + 2\lambda - (x_{2k-1} + x_{2k}) = C_k + \sqrt{2} - (x_{2k-1} + x_{2k})$. Bazza ensures $x_{2k-1}^2 + x_{2k}^2 = 2$. Let $g(t) = t + \sqrt{2 - t^2}$. $C_{k+1} = C_k + \sqrt{2} - g(x_{2k-1})$. Alice chooses $x_{2k-1} \in [0, C_k]$. To ensure Alice survives, we check the minimum possible budget for the next turn. Since $C_k \leq 1/\sqrt{2} < 1$ and $g(t)$ is increasing on $[0, 1]$ (as $g'(t) = 1 - t/\sqrt{2 - t^2} > 0$ for $t < 1$), $g(x_{2k-1})$ is maximized when $x_{2k-1} = C_k$. $C_{k+1} \geq C_k + \sqrt{2} - g(C_k) = \sqrt{2} - \sqrt{2 - C_k^2}$.
    If $C_k > 0$, then $\sqrt{2 - C_k^2} < \sqrt{2}$, so $C_{k+1} > 0$. By induction, $C_k > 0$ for all $k$. Alice survives. The game continues forever. Alice cannot win.
\end{enumerate}

\textbf{Conclusion:} Alice has a winning strategy if and only if $\lambda > \frac{1}{\sqrt{2}}$. Bazza has a winning strategy if and only if $0 < \lambda < \frac{1}{\sqrt{2}}$. If $\lambda = \frac{1}{\sqrt{2}}$, neither player has a winning strategy.
