= Solution
The printed <mean> is $1/(\lambda i)$, so $T_i$ has <exponential distribution> of rate $i\lambda$. Let $S_k=\sum_{i=1}^kT_i$, with fixed $k$. Then
$$
\boxed{\text{speed }N,\qquad I_k^{\rm fixed}(x)=
\begin{cases}\lambda x,&x\geq0,\\+\infty,&x<0.\end{cases}}
$$
Here is a direct proof, which also identifies the exceptional role of the slowest exponential clock. For $0<\theta<\lambda$, <independence> and the <moment-generating function> give
$$
\mathbb E e^{\theta S_k}=\prod_{i=1}^k\frac{i\lambda}{i\lambda-\theta}<\infty,\qquad
\limsup_N N^{-1}\log\mathbb P(S_k\geq Nx)\leq-\theta x.
$$
Letting $\theta\uparrow\lambda$ proves the tail upper bound $-\lambda x$. For a local lower bound at $x>0$, fix $0<\eta<x$ and $M>0$ with $\mathbb P(\sum_{i=2}^kT_i\leq M)>0$; for $k=1$ this <probability> is one. For large $N$, the event
$$
N(x-\eta/2)<T_1<N(x+\eta/2),\qquad
\sum_{i=2}^kT_i\leq M
$$
puts $S_k/N$ in $(x-\eta,x+\eta)$. Its logarithmic <probability> divided by $N$ tends to $-\lambda(x-\eta/2)$. Shrinking $\eta$ gives the local lower bound $-\lambda x$. At zero, $S_k/N\to0$ almost surely and neighborhoods have <probability> tending to one.
For any <closed set> excluding zero, its nonnegative part is bounded away from zero; the tail upper bound applies at its <infimum>. Sets outside the support have <probability> zero. Together with the local lower bounds this proves the full <large deviation principle>, with a <good rate function>. Equivalently, the <product large-deviation principle> for the fixed vector $(T_1/N,\ldots,T_k/N)$ and the <contraction principle for large deviations> minimize $\sum_i i\lambda u_i$ subject to $u_i\geq0$ and $\sum_i u_i=x$; all the excursion is assigned to $u_1$.
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